SolidWorks Motion Tutorials

SolidWorks® 2011

SolidWorks Motion

Dassault Systèmes SolidWorks Corporation300 Baker AvenueConcord, Massachusetts 01742 USA

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© 1995-2010, Dassault Systèmes SolidWorks Corporation, a Dassault Systèmes S.A. company, 300 Baker Avenue, Concord, Mass. 01742 USA. All Rights Reserved.

The information and the software discussed in this document are subject to change without notice and are not commitments by Dassault Systèmes SolidWorks Corporation (DS SolidWorks).

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SolidWorks® 3D mechanical CAD software is protected by U.S. Patents 5,815,154; 6,219,049; 6,219,055; 6,611,725; 6,844,877; 6,898,560; 6,906,712; 7,079,990; 7,477,262; 7,558,705; 7,571,079; 7,590,497; 7,643,027; 7,672,822; 7,688,318; 7,694,238; and foreign patents, (e.g., EP 1,116,190 and JP 3,517,643).

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Document Number: PMT1142-ENG_DRAFT

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Dassault Systèmes SolidWorks Corporation, 300 Baker Avenue, Concord, Massachusetts 01742 USA

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This software is based in part on the work of the Independent JPEG Group.PR

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Contents

Introduction:

About This Course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Prerequisites . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Course Design Philosophy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Using this Book . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Laboratory Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Training Files . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

Windows® 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Conventions Used in this Book . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

Use of Color . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

What is SolidWorks Motion? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

What is Motion Simulation? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Understanding Basics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Mass and Inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Degrees-of- Freedom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Constraining Degrees-of- Freedom . . . . . . . . . . . . . . . . . . . . . . . . 6

Motion analysis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

How is motion analyzed on the computer?. . . . . . . . . . . . . . . . . . . 6

Basics of Mechanism Setup in SolidWorks Motion . . . . . . . . . . . . . . . 7

Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Fixed Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Floating Parts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Mates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Constraint Mapping Concept . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

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Contents SolidWorks 2011

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Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Lesson 1:Introduction toMotion Simulation and Forces

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

Basic Motion Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Case Study: Car Jack Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Driving Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Understanding Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Applied Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Force Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Force Direction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Case 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

Case 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Case 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Results. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Plot Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Sub-Categories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Resizing Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Exercise 1:

3D Fourbar Linkage . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Lesson 2:Building a Motion Model and Post-processing

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Creating Local Mates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

Case Study:

Crank Slider Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34



Mates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Concentric Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Hinge Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Point-to-Point Coincident Mate . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Lock Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

Two Face-to-Face Coincident Mates . . . . . . . . . . . . . . . . . . . . . . 37

Universal Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Screw Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Point-on-Axis Coincident Mate . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Parallel Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

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Perpendicular Mate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

Local Mates. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

Function Builder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Importing Data Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

Power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

Alternative Units. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Plotting Kinematic Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Absolute vs. Relative values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Output coordinate system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Angular Displacement Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Angular Velocity and Acceleration Plots . . . . . . . . . . . . . . . . . . . 62

Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Exercise 2:

Piston . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Exercise 3:

Trace Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Lesson 3:Introduction to Contacts, Springs and Dampers

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Contact and Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

Case Study: Catapult. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76



Interference Detection. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

Contact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Contact groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

Contact Friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

Translational Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Magnitude of Spring Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

Translational Damper . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

Post-processing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

Analysis with Friction (Optional) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

Exercise 4:

The Bug. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

Exercise 5:

Door Closer. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

Lesson 4:Advanced Contact

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

Contact Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

Case Study: Latching Assembly . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

Problem Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

Fixing Motion with Motors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

Motor Input and Force Input Types . . . . . . . . . . . . . . . . . . . . . . 103

Functional Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

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Force Functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

STEP Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

Contact: Solid Bodies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

Poisson Model (Restitution Coefficient) . . . . . . . . . . . . . . . . . . 110

Impact Force Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

Closing Remarks. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

Geometrical Description of Contacts . . . . . . . . . . . . . . . . . . . . . . . . 115

Instability Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

Modifying Result Plots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

Closing Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

Precise Contact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

Integrators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

GSTIFF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

WSTIFF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

SI2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126

Discussion: References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Exercise 6:

Hatchback . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

Exercise 7:

Conveyor Belt (No Friction). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

Path Mate Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144

Exercise 8:

Conveyor Belt (With Friction) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

Lesson 5:Curve to Curve Contact

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153

Contact Forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154

Case Study: Geneva Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . 154


Curve to Curve Contact . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

Solid bodies vs. curve to curve contact. . . . . . . . . . . . . . . . . . . . . . . 159

Solid Bodies Contact Solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

Exercise 9:

Conveyor Belt (Curve to curve contact with friction) . . . . . . . . . . . 161

Lesson 6:CAM Synthesis

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165

CAMs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166

Case Study: CAM Synthesis. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166


Stages in the Process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

Generating a CAM Profile . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

Trace Path . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

Exporting Trace Path Curves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

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Cycle based motion. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

Exercise 10:

Desmodromic CAM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

Exercise 11:

Rocker CAM Profile . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

Lesson 7:Flexible Joints

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 193

Flexible Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194

Case Study:

System with Rigid Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194



Calculation of Wheel Input Motion . . . . . . . . . . . . . . . . . . . . . . 198

Understanding Toe Angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

System with Flexible Joints . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

References. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

Lesson 8:Redundancies

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 209

Redundancies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210

What are redundancies? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213

Effects of Redundancies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214

How are redundancies removed in the solver? . . . . . . . . . . . . . . 215

Case Study:

Door Hinges . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216


Degrees of Freedom Calculation . . . . . . . . . . . . . . . . . . . . . . . . 219

Total Actual and Estimated DOF . . . . . . . . . . . . . . . . . . . . . . . . 219

Using Flexible Joints Option to Remove Redundancies . . . . . . 222

Limitations of Flexible Mates. . . . . . . . . . . . . . . . . . . . . . . . . . . 223

Bushing Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224

How to Check For Redundancies . . . . . . . . . . . . . . . . . . . . . . . . . . . 225

Typical Redundant Mechanisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

Dual Actuators Driving a Part . . . . . . . . . . . . . . . . . . . . . . . . . . 226

Parallel Linkages. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226

Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 227

Exercise 12:

Dynamic Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 229

Exercise 13:

Dynamic Systems 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231

Exercise 14:

Kinematic Mechanism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 233

Exercise 15:

Zero Redundancy Model-Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 238

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Exercise 16:

Zero Redundancy Model-Part 2 (Optional) . . . . . . . . . . . . . . . . . . . 243

Exercise 17:

Removing Redundancies with Bushings . . . . . . . . . . . . . . . . . . . . . 244

Exercise 18:

Catapult . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

Lesson 9:Export to FEA

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

Exporting Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

Case Study: Drive Shaft . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258

Project Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258


FEA Export . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 262

Load Bearing Faces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

Mate location . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

Export of Loads . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264

SolidWorks Simulation Users Only . . . . . . . . . . . . . . . . . . . . . . 266

Direct Solution in SolidWorks Motion . . . . . . . . . . . . . . . . . . . . . . . 274

Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278

Exercise 19:

Export to FEA. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279

Lesson 10:Event Based Simulation

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289

Event Based Simulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290

Case Study: Sorting Device . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290


Servo motors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 290

Sensors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291

Task . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298

Lesson 11:Design Project (Optional)

Objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

Design Project. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 300

Case Study: Surgical Shear - Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . 300


Force to Cut the Catheter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301

Self Guided Problem - Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303


Self Guided Problem - Part 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304


Problem Solution - Part 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 305

Creating the Force Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308

Force to Cut the Catheter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308

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Creating the Force Expression . . . . . . . . . . . . . . . . . . . . . . . . . . 311

Force Expression. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313

IF Statement . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313

Developing the Expression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313

Case Study: Surgical Shear - Part 2 . . . . . . . . . . . . . . . . . . . . . . . . . 320


Summary. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330

Appendix A:Motion Study Convergence Solutions and Advanced Options

Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 334

Accuracy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335

Integrator Type . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

GSTIFF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 336

WSTIFF . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

Stabilized Index Two (SI2). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

Integrator Settings. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

Maximum Iterations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

Initial Integrator Step Size . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337

Minimum Integrator Step Size . . . . . . . . . . . . . . . . . . . . . . . . . . 337

Maximum Integrator Step Size . . . . . . . . . . . . . . . . . . . . . . . . . . 338

Jacobian Re-evaluation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338

Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339

Appendix B:Mate Friction

Mate Friction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342

Concentric (Spherical) Mate Friction Model . . . . . . . . . . . . . . . 343

Coincident Translational Mate Friction Model . . . . . . . . . . . . . 344

Concentric Mate Friction Model. . . . . . . . . . . . . . . . . . . . . . . . . 344

Coincident Mate (Planar) Friction Model. . . . . . . . . . . . . . . . . . 344

Universal Joint Friction Model . . . . . . . . . . . . . . . . . . . . . . . . . . 345

Friction Results Reported . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 345

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Introduction

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Introduction SolidWorks 2011

2

About This Course

The goal of this course is to teach you the basics of how to use the

SolidWorks Motion simulation software to help you analyze the

kinematic or dynamic behavior of your SolidWorks assembly model.

The focus of this course is on the fundamental skills and concepts

central to the successful use of SolidWorks Motion 2011. You should

view the training course manual as a supplement to, and not a

replacement for, the system documentation and on-line help. Once you

have developed a good foundation in basic skills, you can refer to the

on-line help for information on less frequently used command options.

Prerequisites Students attending this course are expected to have the following:

� Mechanical design experience.

� Experience with the Windows™ operating system.

� Completed the on-line SolidWorks tutorials that are available under

Help. You can access the on-line tutorials by clicking Help, Online

Tutorial.

Course Design Philosophy

This course is designed around a process- or task-based approach to

training. Rather than focusing on individual features and functions, a

process-based training course emphasizes processes and procedures

you should follow to complete a particular task. By utilizing case

studies to illustrate these processes, you learn the necessary commands,

options and menus in the context of completing a design task.

Recommended

Length

The recommended minimum length of this course is two days.

Using this Book This training manual is intended to be used in a classroom environment

under the guidance of an experienced SolidWorks Motion instructor. It

is not intended to be a self-paced tutorial. The examples and case

studies are designed to be demonstrated “live” by the instructor.

Please note, there may be slight differences in results for certain lessons

due to service pack upgrades, etc.

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SolidWorks 2011 Introduction

3

Laboratory Exercises

Laboratory exercises give you the opportunity to apply, practice and

expand the material covered during the lecture/demonstration portion

of the course. They are designed to represent typical design, and

analysis situations while being modest enough to be completed during

class time. You should note that many students work at different paces.

Therefore, we have included more lab exercises than you can

reasonably expect to complete during the course. This ensures that even

the fastest student will not run out of exercise.

Training Files A complete set of the various files used throughout this course can be

downloaded from the SolidWorks website, www.solidworks.com.

Click on the link for Support, then Training, then Training Files, then

SolidWorks Simulation Training Files. Select the link for the desired

file set. There may be more than one version of each file set available.

Direct URL:

www.solidworks.com/trainingfilessimulation

The files are supplied in signed, self-extracting executable packages.

The files are organized by lesson number. The Case Studies folder

within each lesson contains the files your instructor uses while

presenting the lessons. The Exercises folder contains any files that are

required for doing the laboratory exercises.

Feature Names Throughout this course, feature names may be different from those you

obtain when doing the case studies and exercises. SolidWorks and

SolidWorks Motion name features sequentially, (Hinge1, Hinge2, etc.)

so if you apply mates in a different order, or you delete and then

recreate a mate, your names will be different. In most cases, mates are

referred to by their type (lock, hinge, coincident, etc.) and the

components that are mated (link, support, etc.). In addition, images are

also included to help avoid ambiguity, but you must always check the

instructions carefully to make sure you are selecting the correct feature.

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Windows® 7 The screen shots in this manual were made using SolidWorks 2010 and

SolidWorks Motion 2010 running on Windows® 7. If you are running

on a different version of Windows, you may notice differences in the

appearance of the menus and windows. These differences do not affect

the performance of the software.

Conventions Used in this Book

This manual uses the following typographic conventions:

Use of Color The SolidWorks and SolidWorks Motion user interface make extensive

use of color to highlight selected geometry and to provide you with

visual feedback. This greatly increases the intuitiveness and ease of use

of the SolidWorks Motion software. To take maximum advantage of

this, the training manuals are printed in full color.

Convention Meaning

Bold Sans Serif SolidWorks Motion commands and options

appear in this style. For example, Motion,

Delete Results means choose the Delete

Results option from the Motion menu.

Typewriter Feature names and file names appear in this

style. For example, Concentric.

17 Do this step

Double lines precede and follow sections of

the procedures. This provides separation

between the steps of the procedure and large

blocks of explanatory text. The steps

themselves are numbered in sans serif bold.

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What is SolidWorks

Motion?

SolidWorks Motion is a virtual prototyping tool for engineers and

designers interested in understanding the performance of their

assemblies. Powered by ADAMS® technology, the industry leader for

over 25 years, SolidWorks Motion helps you ensure that your designs

will work and perform as expected prior to building them. By learning

how to effectively utilize the features of the user interface, you will

have the key to unlocking a solution to the most complex mechanisms.

What is Motion Simulation?

A mechanism is a mechanical device that has the purpose of

transferring motion and/or force from a source to an output. Motion

simulation is simply the study of such moving systems or mechanisms.

The motion of any system is determined by the following:

� Mates connecting the parts

� The mass and inertia properties of the components

� Applied forces to the system

� Driving motions (Motors or Actuators)

� Time

Understanding Basics

Mass and Inertia The principle of inertia is one of the fundamental laws of classical

physics which are used to describe the motion of matter and how it is

affected by applied forces. Today, the concept of inertia is most

commonly defined using Isaac Newton's First Law of Motion, which

states:

Every body perseveres in its state of being at rest or of moving

uniformly straight ahead, except insofar as it is compelled to

change its state by forces impressed.

Mass and inertia play a very important role in the simulation of

dynamic systems and are also important in kinematics. Realistic values

of mass and inertia are needed in nearly all simulations.

Degrees-of- Freedom

An unconstrained rigid body in

space has six degrees-of-freedom:

three translational and three

rotational. It can translate along its

X, Y, and Z axes and rotate about its

X, Y, and Z axes as shown in the

figure to the right.

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Constraining Degrees-of- Freedom

Constraints are the

restrictions placed on

a part’s movement in

specific degrees-of-

freedom. Mates are

connections that

restrict the movement

of one part with

respect to another.

Motion analysis The two equations governing three dimensional motion of a rigid body

are known as Euler’s equations.

The first equation is Newton’s second law of motion which states that

the sum of externally applied forces on a body is equal to the rate of

change of linear momentum P, .

For bodies where mass does not change, the right hand side of the

equation simplifies to more commonly known mass times acceleration,

.

The second equation is based on the sum of the moments about the

center of mass of a rigid body due to external forces, and couples

should equal the rate of change of angular momentum H of the body.

How is motion analyzed on the computer?

Τhe program uses the modified Newton Raphson iteration method in

each time step.

By taking very small time steps, the software can predict the position of

parts at the next time step based on initial conditions or the previous

time step.

The solution must satisfy:

� Velocity of parts

� Mates connecting parts

� Forces and accelerations

ΣFdP

dt--------=

ΣF ma=

ΣMdH

dt--------=

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The answer is iterated until certain accuracy is reached for that time

step for force and acceleration values.

Basics of Mechanism Setup in SolidWorks

Motion

The following paragraphs outline how SolidWorks Motion treats parts

and sub-assemblies, and how the mates directly define the motion of

the mechanism when loaded by external forces (such as gravity or

isolated forces) or prescribed motions (motors).

Rigid Body In SolidWorks Motion, all parts are treated as infinitely rigid. This

means that there is no internal deflection within a part and the parts do

not deform or change shape during the simulation. A rigid body can be

a single SolidWorks part or a sub-assembly.

There are two states of the sub-assembly in SolidWorks: Rigid or

Flexible. A rigid sub-assembly means that the individual components

that make up the sub-assembly are assumed to be rigidly attached

(welded) to each other as if they were one single part.

If a sub-assembly status is set to flexible in SolidWorks, it does not

mean that the sub-assembly parts become flexible. This setting means

that the root level parts of the sub-assembly are treated independently

of each other by SolidWorks Motion. The constraints (SolidWorks

mates at the sub-assembly level) between these parts are automatically

mapped into SolidWorks Motion.

Fixed Parts A rigid body can be treated as a fixed part or a floating (moving) part.

Fixed parts are, by definition, at absolute rest. Each fixed rigid body

has zero degrees-of-freedom. A fixed part serves as the reference frame

for the remaining rigid bodies that are moving.

In SolidWorks, any component that is fixed in your assembly is

automatically treated as a fixed part when you begin a new mechanism

and map the assembly constraints.

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Floating Parts Components that move in the mechanism are considered moving parts.

Each moving part has six degrees-of-freedom.

In SolidWorks, any component that is floating in your assembly is

automatically treated as moving when you begin a new mechanism and

map the assembly constraints.

Mates SolidWorks mates fully define how rigid bodies are attached and how

they move relative to each other. Mates remove degrees-of-freedom

from the parts to which they are attached.

When you add a mate, such as a concentric mate, between two rigid

bodies, you remove degrees-of-freedom, causing them to remain

positioned with respect to each other regardless of any motion or force

in the mechanism.

Motors Motors can be defined for part to control its movement over a period of

time. A motor dictates the displacement, velocity, or acceleration of a

part as a function of time.

Gravity Gravity is an important quantity when the weight of a part has an

influence on its simulated motion, such as a body in free fall. In

SolidWorks Motion, gravity consists of two components:

� Direction of the gravitational vector

� Magnitude of gravitational acceleration

The Gravity Properties dialog allows you to specify the direction and

magnitude of the gravitational vector. You can specify the gravitational

vector by entering the x, y and z values in the appropriate text box, or

by specifying a reference plane. The magnitude must be entered

separately. The default value for the gravitational vector is (0, -1, 0),

and the magnitude is 9.81 m/s2 (or the equivalent in the currently active

units).

Constraint Mapping Concept

One of the reasons SolidWorks Motion is such a time saving tool is that

it automatically maps the SolidWorks assembly mates (constraints) to

SolidWorks Motion. There are more than 100 ways to mate or constrain

parts in SolidWorks.

Forces When defining various Force objects in SolidWorks Motion, a location

and/or direction has to be specified. These directions and locations are

derived from selected SolidWorks entities. The entities can be sketch

points, vertices, edges or surfaces.

Summary This short review of motion simulation using SolidWorks Motion is

not, of course, all inclusive. It is only intended to get us started with the

hands-on lessons. As we progress through the lessons presented in the

following chapters, we will occasionally digress from software-specific

issues in order to discuss relevant motion simulation fundamentals.

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Lesson 1Introduction to

Motion Simulationand Forces

Objectives Upon successful completion of this lesson, you will be able to:

� Use Assembly Motion to animate the motion of a car jack

assembly.

� Use SolidWorks Motion to simulate physical behavior of the car

jack and determine the torque required to lift a vehicle.

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Lesson 1 SolidWorks 2011Introduction to Motion Simulation and Forces

10

Basic Motion Analysis

In this lesson, we will perform a basic motion analysis using

SolidWorks Motion to simulate the weight of a vehicle on the jack and

determine the torque required to lift it. Engineers can then use this

information to choose the required electric motor to drive the car jack.

Case Study: Car Jack Analysis

A mechanical jack is a device that lifts heavy equipment. The most

common form is a car jack, floor jack, or garage jack which lifts

vehicles so that maintenance can be performed. Car jacks usually use

mechanical advantage to allow a human to lift a vehicle. More

powerful jacks use hydraulic power to provide more lift over greater

distances. Mechanical jacks are usually rated for a maximum lifting

capacity (e.g., 1.5 tons or 3 tons).

Because this is our first motion analysis, no contact is used and the

tilting motion of the jack is prevented with the help of the mates.

Problem Description

The car jack will be driven at a rate of 100 RPM and will be loaded

with a force of 8,900 N., representing the weight of a vehicle.

Determine the torque and power required to lift the load through the

range of motion of the jack.

Stages in the Process

� Create a Motion Study.

This will be a new motion study.

� Add a rotary motor.

The rotary motor will drive the jack.

� Add gravity.

Normal gravity will be added so that the weight of the car jack

components are considered in the calculations.

� Add the weight of the car.

The weight of the car will be added as a downward force on the

Support.

� Calculate the motion.

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SolidWorks 2011 Lesson 1Introduction to Motion Simulation and Forces

11

The default analysis will run for five seconds but we will increase it

to allow the jack to extend fully.

� Plot the results.

We will create various plots to show the torque and power required.

1 Ensure that SolidWorks Motion is

added in.

Under Tools, Add-ins, make sure

SolidWorks Motion is checked.

2 Open an assembly file.

Open Car_Jack from the

Lesson01\Case Studies folder.

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3 Set the document units.

SolidWorks Motion uses the document units set in the SolidWorks

document.

Click Tools, Options, Document Properties, Units.

Select MMGS (millimeter, gram, second) for the Unit system. This

will set our length units to millimeters and force to Newtons.

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4 Change to the Motion Study.

Click on the Motion Study 1 tab that appears at the bottom left-hand

corner of the window. If this tab is not visible, select Motion Manager

on the View menu.

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Driving Motion Motion can be driven by gravity, springs, forces or motors. Each has

different characteristics that can be controlled.

Introducing: Motors Motors can create either linear, rotary or path dependent motion or to

prevent motion. This motion can be defined in a number of different

ways.

� Constant Speed

The motor will drive at a constant velocity.

� Distance

The motor will move for a fixed distance or degrees.

� Oscillating

Oscillating motion is a back and forth motion at a specific distance

at a specified frequency.

� Segments

Motion profile is constructed from segments of the most commonly

used functions such as linear, polynomial, half-sine and others.

� Data Points

Interpolated motion is driven by a tabular set of values.

� Expression

The motor can be driven by a function created from existing

variables and constants.

� Servo Motor

The motor used to implement control actions for the event-based

triggered motion.

Where to Find It � On the MotionManager toolbar, click Motor .

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5 Create a Motor that drives the Screw_rod at 100 RPM.

Click Motor on the Motion Manager toolbar.

Under Motor Type, select Rotary Motor.

Under Component Direction, select the cylindrical face of the

Screw_rod part as shown in the figure. The Motion Direction field

will automatically populate the same face to specify the direction.

Use the Reverse Direction button to orient the motor (see the figure).

Leave the Component to move relative to field empty. This ensures

that the motor direction is specified with respect to the global

coordinate system.

Under Motion, select the Constant speed and enter a value of

100 RPM.

Click OK.

Important! Make sure that the motor is oriented as shown in the figure.

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Click the graph in the PropertyManager to view the enlarged plot.

Close the graph plot and click OK to close the Motor PropertyManager.

6 Type of Study.

Make sure that the Motion Type of Study field

shows Motion Analysis.

Gravity Gravity is an important quantity when the weight of a part has an

influence on its simulated motion, such as a body in free fall. In

SolidWorks Motion, gravity consists of two components:

� Direction of the gravitational vector

� Magnitude of the gravitational acceleration

The Gravity Properties allows you to specify the direction and

magnitude of the gravitational vector. You can specify the gravitational

vector by selecting the X, Y and Z direction or by specifying a

reference plane. The magnitude must be entered separately. The default

value for the gravitational vector is Y and the magnitude is

9806.55 mm/sec2 or the equivalent in the currently active units.

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7 Apply Gravity to the assembly.

Click Gravity on the Motion Manager toolbar.

For Gravity Parameters, Direction Reference,

select the Y direction.

Under Numeric gravity value, type in a value of

9806.65 mm/sec^2.

Click OK.

Forces Force entities (including both forces and moments) are used to effect

the dynamic behavior of parts and sub assemblies of a motion model

and are usually a representation of some external effect acting on the

analyzed assembly.

Forces may resist or induce motion, and are defined using similar

functions that are used to define motors (constant, step, function,

expression or interpolated).

Forces in SolidWorks Motion can be divided into two basic groups:

� Action Forces

A single applied force or moment

representing the effect of the external

objects and loadings on the part or

subassembly. The weight of the vehicle

applied on the car jack or an

aerodynamic force on the car body are examples of action forces.

� Action and Reaction Forces

A pair of forces or moments, both action and corresponding

reaction, are applied on the parts or subassemblies.

A spring force could be understood as action and reaction force

because both are acting on the same line of action and acting on the

assembly at the spring mount points. Another example would be a

person pushing with his/her arms on the two opposing parts of an

assembly. Such a person can then be represented in the motion

analysis by a pair of two opposing forces of equal magnitude on the

same line of action, i.e. action and reaction forces.

Understanding Forces

A force can define load or compliance on a part. SolidWorks Motion

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Applied Forces Applied forces are forces that define loads at specific locations on a

part. You must provide you own description of the force behavior by

specifying a constant force value or a function expression. The applied

forces available in SolidWorks Motion are the applied force, applied

torque, action/reaction force and action/reaction torque.

The orientation of action-only forces can be fixed or at relative to the

orientation of any part in the mechanism.

Applied forces are used to model inputs such as actuators, rockets,

aerodynamic loads and many more.

Force Definition To define a force the following information must be specified:

� Part or parts on which the forces act.

� Point of the force application.

� Magnitude and direction of the force.

Where to Find It � On the MotionManager toolbar, click Force . Select Action-

Only in the PropertyManager.

Force Direction The force direction is based on the reference part

you select in the Force Direction box. An

illustration below gives you the three cases on how

the force direction changes based on the selected

reference parts.

Case 1 Direction of force is based on a fixed component.

If fixed component is the assembly origin then the initial orientation of

the force will be held constant throughout the simulation.

F1

Reference Fixed Component

F1

F1

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Case 2 Direction of force is based on the selected moving component,

which is also the component on which you want to apply the

force.

If the part to which the force is applied is used as the reference datum,

then the force will remain locked in its relative orientation to the body

over the entire simulation time (i.e. it will stay in alignment with the

geometry on the body used to define the direction).

Case 3 Direction of force is based on the selected moving component

which is different from the component on which you want to

apply the force.

If another moving part is used as the reference datum, the direction of

the force will change based on the relative orientation of the reference

body to the moving body. This is hard to visualize easily, but if you

apply the force on a body that is held locked in position, and use a

rotating part as the reference datum, you should see the force rotates in

concert with the reference body.

Note Make sure that the gravity symbol shows the orientation in the

negative Y direction.

F1

Reference Rotating Component

F1

F1

Fixed Component

F1

Reference Rotating Component

F1F1

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8 Create a force of 8900 N to simulate the weight of the car on the

car jack.

Click Force on the Motion Manager.

For Type, select Force.

Under Direction, select Action Only.

Under Action part and point of application of action, select the

circular edge on component Support-1 (see image below).

For Force Direction, select the vertical edge on the Base-1

component.

Note The default force direction is defined by the circular edge selected in

the Action part and point of application of action field, i.e.

perpendicular to the plane of the edge. Because the default direction is

correct in this case, the edge selected in the Force Direction field is not

required and is done solely for the educational purpose.

Under Force Function, select Constant. Enter a force value of

8900 N.

Note Make sure that the force is directed downwards.

Click OK to close the Force/Torque PropertyManager.

9 Run the Simulation.

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10 Run the Simulation for 8 seconds.

Drag the end time key to 8 seconds on the timeline and recalculate.

Results The primary output from a motion study is a plot of one parameter

versus another, usually time.

Once the motion is calculated plots can be created for a variety of

parameters. All existing plots will be listed at the bottom of the

MotionManager tree.

Plot Categories Plots of the following categories can be created:

� Displacement � Displacement

� Acceleration � Forces

� Momentum � Energy

� Power � Other quantities

Sub-Categories Within each of the categories, plots can be created for:

� Trace Path � XYZ Position

� Linear Displacement � Linear Velocity

� Linear Acceleration � Angular Displacement

� Angular Velocity � Angular Acceleration

� Applied Force � Applied Torque

� Reaction Force � Reaction Moment

� Friction Force � Friction Moment

� Contact Force � Translational Momentum

� Angular Momentum � Translational Kinetic Energy

� Angular Kinetic Energy � Total Kinetic Energy

� Potential Energy Delta � Power Consumption

� Pitch � Yaw

� Roll � Rodriguez Parameters

� Bryant Angles � Projection Angles

Resizing Plots Plots can be resized by dragging any border or corner.

Where to Find It � Click Results and Plots on the MotionManager toolbar.

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11 Plot the torque required to lift the weight of the car.

Click Results and Plots in the Motion manager.

Under Result, select the category as Forces.

Under Sub-category, select Motor Torque.

Under Result component, select Magnitude.

Under Select rotational motor object to create result, select the

motor that we created (see image below).

Click OK.

The plot of torque required appears in the graphics area.

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Note Once the Rotary Motor1 is selected, a triad is displayed in the

graphics area. This triad indicates the local X, Y and Z axes of the

motor in which the output quantities may be displayed. In the present

case we require the plot of the magnitude which is independent of the

coordinate system. The post-processing is described in greater detail in

the next lesson.

12 Plot the power consumed to lift a weight of 8900 N.

We will add this plot into an existing graph. Click Results and Plots in

the Motion Manager toolbar.

Under Result, select the category as Momentum/Energy/Power.

Under Sub-category, select Power Consumption.

Under Select motor object to create result, select the same motor

that you selected in the previous step.

Under Plot Results, select Add to existing plot and select Plot1 from

the pull down menu.

Click OK.

The power consumption is 76 Watts. Based on the torque and the power

information, we can select an electric motor and use it to drive the

Screw_rod instead of a human hand.

You can click Play to see the animation. The vertical time bar in

both the MotionManager and the graph indicates the time.

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13 Plot the vertical position of the Support.

Click on the Results and Plots icon in the Motion Manager.

Under Result, select the category as Displacement/Velocity/

Acceleration.

Under Sub-category, select Linear Displacement.

For Result Component, select Y-component.

For Select two points/faces, select the top face of the support. If no

second item is selected, the ground serves as the default second

component, or the reference.

Leave the Component to define XYZ directions field empty. This

indicates that the displacement is reported in the default global

coordinate system.

Note The displacement is measured at the origin of the Support part file,

indicated as the small blue sphere in the above figure, with respect to

the origin of the Car_Jack assembly file. The result is reported in the

default global coordinate system.

Click OK.

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The above graph indicates change of the global Y coordinate of the

origin of the Support part file. The displacement is therefore 51mm

(212-161mm) in the positive global Y axis.

14 Modify the graph.

Modify the ordinate of the graph to show the

angular displacement of the motor.

In the MotionManager tree, expand the Results

folder. Right-click Plot2 and click Edit Feature.

Under Plot Results, Plot Results verus: select

New Result.

For Define new result, select Displacement/

Velocity/Acceleration.

Select Angular Displacement under sub-category.

Select Magnitude for result component.

Select RotaryMotor1 for the simulation element.

Click OK.

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15 Examine the graph.

The result plot is a little coarse and the data ordinate does not cover the

full range of -180 to 180 degrees. To generate a graph with finer detail,

more data must be saved to disk.

Introducing: Study

Properties

SolidWorks Motion has its own set of properties to control the way the

study is calculated and displayed.

Study properties will be discussed throughout the book.

Where to Find It � Click Motion Study Properties on the MotionManager toolbar.

Introducing: Frames

per Second

Frames per second controls how often the data is saved on the disk. The

higher the frames per second, the more dense the data recorded.

Where to Find It � In the Motion Study Properties, expand Motion Analysis and either

type the number, use the spinbox arrows or adjust the slider.

16 Modify Motion Study properties.

Click Motion Study Properties in the

MotionManager toolbar.

Change the Frames per second to 100.

Click OK.

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17 Calculate the study.

Notice that we have more detail and the angular displacement is nearly

from -180 to 180 degrees.

18 Save and close the file.

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SolidWorks 2011 Exercise 13D Fourbar Linkage

29

Exercise 1:3D Fourbar Linkage

This assembly is a simple mechanism called 3D Fourbar linkage.

There are only four parts in the mechanism. The Support part is

grounded, and the rotation of the Lever part will cause a sliding motion

of the SliderBlock part.

This exercise reinforces the following skills:

� Basic Motion Analysis on page 10.

� Results on page 21.

Project Description

The LeverArm will be simply rotated with a constant 360 deg/sec

angular velocity. Determine the amount of torque required to drive this

mechanism and plot it from the motion simulation.


Open 3D Fourbar linkage from the Lesson01\Exercises folder.

2 Verify fixed and moving components.

Make sure that support is fixed while the other

components can move.

3 Motion study.

In the MotionManager, select Motion Analysis.

The default Motion Study 1 will be used for the analysis.

4 Add gravity.

Apply gravity in the negative Z direction.

LeverArm

linkage

SliderBlock

Support

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Exercise 1 SolidWorks 20113D Fourbar Linkage

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5 Define motion of the Lever Arm.

Define a Rotary Motor at 360 deg/sec.

Tip You can enter 360 deg/sec directly into the PropertyManager and it will

automatically be converted to RPM.

6 Motion study properties.

Set the Frames per second to 100 and drag the time key to 4

seconds.

7 Calculate the simulation.

8 Determine the torque and power required to drive the mechanism.

Define a graph showing the moment torque and the required power as a

function of time. Define both quantities in a single graph.

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SolidWorks 2011 Exercise 13D Fourbar Linkage

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9 Linear velocity of the SliderBlock.

Plot a graph showing the linear velocity of the SliderBlock as a

function of time.

10 Modify the graph.

Modify the ordinate of the graph to show the angular displacement of

the Rotary Motor. This way the graph will show the variation of the

SliderBlock velocity relative to the angular displacement of the

LeverArm.


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Exercise 1 SolidWorks 20113D Fourbar Linkage

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33

Lesson 2Building a Motion Model

and Post-processing


� Build proper SolidWorks Motion models for kinematic simulation.

� Create local mates for a SolidWorks Motion study.

� Create and modify plots for post-processing.

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Lesson 2 SolidWorks 2011Building a Motion Model and Post-processing

34

Creating Local Mates

In the previous lesson, the mates created in SolidWorks were used as

joints in SolidWorks Motion. If the components are not mated in

SolidWorks, or if we wish to examine different connection types in

SolidWorks Motion, mates can be added or modified in the Motion

Analysis.

Case Study:Crank Slider Analysis

In this lesson, we will setup the mechanism for the crank slider model.

We will use SolidWorks mates that most closely represent the real

mechanical connections. The crank slider model is used in a variety of

engineering applications, such as a steam engine or the cylinder of an

internal combustion engine. Therefore, we will apply a motor on the

crank part, run the simulation, and then postprocess some results to

estimate the required torque.

Problem Description

The crank is driven at a constant angular velocity of 60 RPM.

Determine the torque required to rotate the crank part.


� Create a motion study.

� Preprocessing.

Add local mates to the assembly with the motion study active.

� Run the simulation.

Calculate the motion.

� Post-processing.

Plot and analyze the results.


Open 3dcrankslider from the Lesson02\Case Studies folder.

Arm Mount

Link1

Crank

Crank Arm

Collar Shaft

Collar

Link2

Housing

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SolidWorks 2011 Lesson 2Building a Motion Model and Post-processing

35

2 Examine the assembly.

SolidWorks Motion assumes that all components

that are fixed in SolidWorks are considered to be

grounded parts, and all components that are

floating are assumed to be moving parts.

However, the movement of these parts is

constrained by the SolidWorks mates.

There are no mates in this assembly, but three

parts are fixed. The collar_shaft, arm_mount and crank_housing are fixed as these are parts

that would be connect to ground and will have no

motion in the assembly.

The remaining parts will need mates to constrain

their motion to that expected of the mechanical system.

Mates Mates are used to constrain the relative motion of a pair of rigid bodies

by physically connecting them.

Note A rigid body acts and moves as a single unit. SolidWorks components

situated at the root level are considered rigid bodies. This means that

SolidWorks and SolidWorks Motion treat subassemblies as single rigid

bodies.

Mates can be classified into two main types:

� Mates used to constrain the relative motion of a pair of rigid bodies

by physically connecting them. Examples: Hinge, Concentric,

Coincident, Fixed, Screw, Cam, etc.

� Mates used to enforce standard geometric constraints. Examples:

Distance, Angle, Parallel, etc.

Below are some descriptions of some of the most commonly used mate

types. For a comprehensive understanding of all the other mates, please

refer to the SolidWorks help.

No

Fixed

Mates

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Concentric Mate The concentric mate allows both relative rotation as well as relative

translation of one rigid body with respect to another rigid body. The

concentric mate origin can be located anywhere along the axis about

which the rigid bodies can rotate or slide with respect to each other.

Example: Piston sliding and rotating inside a cylinder.

Hinge Mate Hinge mate is essentially concentric mate with the restricted translation

between the two components.

In SolidWorks Motion, the hinge mate is used rather than a

combination of concentric and coincident because the mechanical joint

is a hinge. Hinge mates are found in the Mechanical Mates tab of the

Mate PropertyManager.

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Point-to-Point Coincident Mate

This type of mate permits free rotation about a common point of one

rigid body with respect to another rigid body. The origin location of this

mate determines the point about which the rigid bodies can pivot freely

with respect to each other. Example: Ball and Socket joint.

Lock Mate The lock mate locks two rigid bodies together so they may not move

with respect to each other. For a lock mate, the origin location and

orientation does not affect the outcome of the simulation. A real world

example of a lock mate is a weld that holds two parts together.

Two Face-to-Face Coincident Mates

This mate allows one rigid body to translate along a vector with respect

to a second rigid body. The rigid bodies may only translate, not rotate,

with respect to each other.

The location of the origin of a translational joint with respect to its rigid

bodies does not affect the motion of the two bodies but does affect the

reaction or the bearing loads.

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Universal Mate A universal mate permits the transfer of rotation from one rigid body to

another rigid body. This mate is particularly useful to transfer rotational

motion around corners, or to transfer rotational motion between two

connected shafts that are permitted to bend at the connection point

(such as the drive shaft on an automobile).

The origin location of the universal mate represents the connection

point of the two rigid bodies. The two shaft axes identify the center

lines of the two rigid bodies connected by the universal joint. Note that

SolidWorks Motion uses rotational axes parallel to the rotational axes

you identify but passing through the origin of the universal mate.

Screw Mate The screw mate constrains one rigid body to rotate as it translates with

respect to another rigid body.

When defining a screw mate, you can define the pitch. The pitch is the

amount of relative translational displacement between the rigid bodies

for each full rotation of the first rigid body. The displacement of the

first rigid body relative to the second rigid body is a function of the

rotation of the first rigid body about the axis of rotation. For every full

rotation, the displacement of the first rigid body along the translation

axis with respect to the second rigid body is equal to the value of the

pitch.

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Point-on-Axis Coincident Mate

This type of mate permits one translational and three rotational motions

of one part with respect to another. The translational motion between

the parts is confined to the orientation axis. The point defines the initial

pivot location on the axis.

Parallel Mate A parallel mate permits only translational motion of one part with

respect to another. No rotation is allowed.

In the picture below, the blue x part can move relative to the ground in

the X direction. The red y part can move relative to the x part in the Y

direction. The z part can move relative to the y part in the Z direction.

Finally, the red/yellow/blue cube on the z part has a curvilinear motion

relative to the ground but always stays parallel.

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Perpendicular Mate

The perpendicular mate allows both translational and rotational motion

of one part with respect to another. It imposes a single rotational

constraint on the components so that the component axes remain

perpendicular. This allows relative rotations about either z-axis, but

does not allow any relative rotation in the direction perpendicular to

both z-axes.

It is recommended that the mates are representing the real mechanical

connections as closely as possible, i.e. mechanical hinge should be

modeled using the hinge mate and not using a combination of

coincident and concentric mates.

Local Mates Mates created in SolidWorks can be transferred to the Motion Analysis

and used as mechanical joints. If there are no mates in the SolidWorks

assembly or if we wish to define the connections differently than the

SolidWorks mates, we can add local mates directly to the motion study.

Local mates only apply to the study to which they were added.

To add local mates, make a motion study active and add the mates.

With a motion study active, any mate added is only applied in that

motion study.

3 Verify the document units.

Verify that the document units are set to MMGS (millimeter, gram,

second).

4 Create a Motion Study.

Right-click the Motion Study 1 tab and click Create New Motion

Study.

Make sure that the Motion Analysis is selected as the Type of Study

in the MotionManager.

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5 Move components.

Move the components that are not fixed to separate the assembly. We

are doing this only to make it easier to select faces and to keep track of

what components are mated.

6 Create a local mate.

Add a mate and select Hinge from the Mechanical Mates section. For

Concentric Selections, select the two cylindrical faces of the shaft

and hole shown with red arrows. For Coincident Selections, select the

end face of the shaft and crank housing shown with the blue arrows.

Click OK.

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7 Warning.

Because the timeline is active, the mate changes the position of the

crank at the starting position of the animation. This is OK for what we

are doing.

Click Yes.

8 Examine the mate.

Notice that this mate is only located in the MotionManager and not in

the FeatureManager design tree.

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9 Add additional mates.

Add a concentric mate between the two

spherical surfaces shown on the parts

Link1 and crank.

10 Mate arm to arm_mount.

Add a hinge mate to connect the arm to the

arm_mount.

11 Mate Link1 to the arm.

This connection requires two hinge mates, one between Link1 and

cardian, and a second hinge mate between cardian and arm.

Concentric

Link1

crank

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12 Mate Link2.

Link2 will use a hinge mate to

connect the arm. As there is no pin

going through the two holes, the

coincident selections will be the two

touching faces.

Mate the other end of Link2 to the

pin on the collar with just a

concentric mate.

13 Mate collar to collar_shaft.

Add a concentric mate between a cylindrical surface on each part.

14 Test the assembly.

Rotate the crank and make sure the components move as expected.

Check the FeatureManager design tree and the Motion Study tree. All

the mates should just be in the Motion Study tree.

15 Add gravity.

Add gravity in the negative Y direction.

16 Calculate.

Adjust the assembly key to 5 seconds. Click Calculate Simulation .

17 Play the simulation.

Play the simulation at 25% speed.

The crank will rock back and forth as gravity affects the components

and potential and kinetic energy are exchanged. As there is no friction,

the parts will continue to move without end.

18 Set the timebar to 0s.

To add a motor at time 0s, the timebar needs to be set to 0s.

19 Add a motor.

Create a Motor that drives the crank.

Click Motor on the Motion Manager.


Under Motor Location, select the cylindrical

face of the crank part (as shown in the

figure).

The default selection for the Motor

Direction is correct for this analysis. Make

sure that the motor is oriented as shown in

the figure.

Hinge

Concentric

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Under Motion, select the Motor Type as Data Points. The command

invokes the Function Builder window.

Make sure that Value (y) and Independent variable (x) are set to

Displacements (deg) and Time (s).

Function Builder Function Builder can be used to construct functional expression for

motors and forces.

Introducing:

Function Builder

Function Builder can build functional expressions using predefined

Segments, imported set of discrete Data Points or mathematical

Expressions.

The figure below shows the segment view of the Function Builder

window.

� Segments

In Segment view, user select both the independent (typically time)

and dependent variable (displacements, velocity or acceleration).

For each specified interval, the transition from the initial to final

value is controlled using one of the predefined profiles curves. The

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Quarter-Sine, Half-Sine, 3-4-5 Polynomial and others. As the

function is constructed, the graph windows show the corresponding

variation of displacement, velocity, acceleration and the jerk (time

derivative of acceleration). Note that it is possible to save and

retrieve function from stored location.

� Data Points

The discrete set of data points can be imported from a *.csv file or

entered manually. The functionality as well as the options are

similar to the Interpolated option of the input and are explained in

this lesson.

� Expression

Expression enables the construction of functions with the help of

predefined mathematical functions, variables and constants, and

existing motion study results. As in both previous cases, the

function can be saved at a specific location. This procedure will be

used in this lesson.

Where to Find It � In the Motor or Force/Torque PropertyManagers, under Motor

Type or Force Function dialog select Segments, Data Points or

Expression.

20 Import data points.

Rather than type the individual values into the table, we can load them

from a file. In this case, we have an Excel file. Locate the file crank rotation.csv in the Case Studies folder and examine the file. It is just

two columns of numbers representing the time and displacement.

Click the Import Data button. Navigate to and select the crank rotation.csv file and click Open. The values from the file are now

inserted into the Time and Value columns.

Select Akima as the Interpolation type.

Note The Function Builder graph windows automatically updates the plots

for displacement, velocity, acceleration and jerk. The data points

describe linear increase of the angular displacement in time, a harmonic

motion

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.

Click OK to complete the definition of the profile and close the

Function Builder.

Click OK to complete the definition of the Motor feature.

Rename this motor feature to Motor - crank.

Importing Data Points

Using imported data points, you can use your own motion data to

control the displacement, velocity, or acceleration of the motion. The

data points that can be imported into SolidWorks Motion must be in a

text file (*.txt) or comma separated file (*.csv) format. The file should

contain one data point per line. The data point consists of two values,

the time and the value at that time. Commas or spaces can be used as

separators between the values. The file is essentially free format aside

from these restriction. SolidWorks Motion allows for unlimited number

of data points to be used. The minimum number of data points to be

defined is four.

The first column, Independent variable (x), in the data point template

is typically time, but other parameters such as cycle angle, angular

displacement and others can be used as well. The second column,

Value (y), is the displacement, velocity, or acceleration. These values

can be manually entered or imported.

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Besides Linear interpolation, two spline-fitting options are available to

smoothen out the data: the Akima spline (AKISPL) and cubic curve

spline (CUBSPL). It is recommended that you use a cubic curve

because it will work well even if you data points are not evenly spaced.

An Akima curve is fast, but will not work as well if you points are not

evenly spaced.

21 Run the simulation.

Click Calculate to run the simulation for 5 seconds.

22 Plot the torque.

Create a plot for the torque required to turn the

mechanism.

Define the plot by Forces, Motor Torque and

Magnitude.

Select the Motor-crank for the Simulation

element.

Click OK.

23 Examine the plot.

The plot may be improved by recording more data points by increasing

the Frames per second option in the Motion Study Properties.

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24 Plot the power.

Create a plot for the power required to turn the

mechanism.

Define the plot by Momentum/Energy/Power,

and Power Consumption.

Select the Motor-crank for the Simulation

element.

Click OK.

Note Knowing the operating RPM, torque and/or power we can select the

appropriate motor to drive our system.

Power Power is the rate at which work is performed, or the amount of work

conducted in one second. Forces conduct work on distances, moments

then on the angular displacements. For rotating motors the following

relationship therefore holds:

The power plot in the previous figure can be easily verified. The

maximum torque is 10 N-mm = 0.01 N-m

Students can easily verify this by creating the plot of the angular

velocity.

The resulting maximum power is then:

The graph of the power indicates 0.06 W because two significant digits

precision is used by default.

Power [W] Torque [N-m] Angular velocity [rad/sec]×=

Angular velocity 360 deg/sec 2π rad/sec= =

Power 0.01 2× π 0.063W= =PRE-

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Alternative Units Often times the rating of the electric motors is expressed in maximum

power and torque. Alternative units are used frequently.

If rpm is used for the angular velocity, then:

If horsepower is used instead, the following conversion can be used:

A useful formula when computing power using mechanical horsepower

in the English system of units is the following:

While mechanical horsepower is common in some industries in the

United States (automotive industry, for example), similar measure

called metric horsepower is used in Europe and Asia. Metric

horsepower is then defined as:

Because of this ambiguity in the definition of horsepower, its use today

is not recommended.

25 Add a mate.

When we added mates to this motion study, we only added mates

essential to describing the motion. Depending on how the assembly is

built, a mate preventing the collar from rotating around the collar shaft

could be defined. This mate would represent the mechanical function of

the keyway.

Add a coincident mate between one side face of the key and the

corresponding face on the keyway.

Power [W]Torque [N-m] 2π× Angular velocity [rpm]×

60-------------------------------------------------------------------------------------------------------------------------=

Mechanical horsepower 33,000 lb-ft/min 745.7W= =

Power [hp]Torque [lb-ft] 2π× Angular velocity [rpm]×

33,000------------------------------------------------------------------------------------------------------

Torque[lb-ft] RPM×5252.1

-----------------------------------------------= =

Metric horsepower 735.5W=

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26 Run the study.

Change the Frames per Second to 100, then re-calculate the study.

27 Review results.

The torque values are essentially the same as when we did not have the

coincident mate. The plot is now smoother as we have four times more

data points.

Following the recommendations that all mates should represent the real

mechanical connections for the kinematic analyses, this mate defining

the keyway could be defined, even if it is not required for the actual

motion analysis.

28 Plot reaction force.

Create a new plot to show the reaction force on

the motor.

Define the plot by Forces, Reaction Force, and

Magnitude.

In this assembly, the first hinge we defined was

between the crank and crank_housing. As the

crank_housing is fixed, the mate must transmit

the reaction force.

Select the first hinge mate as the Simulation

element.

Because the selected mate connects two parts,

there are two equal and opposite forces acting in

the mate. One of the two parts must be selected

for the plot of this force.

Select any face on the crank-1 part as the second component in the

Simulation Element field.

Click the Show vector in the graphics window checkbox.

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Note This mate must be selected from the Mate Group 1 folder in the

Motion Manager tree because this mate is local and is not listed in the

FeatureManager design tree.

Click OK.

29 Warning.

We will receive a warning about redundant constraints. Redundant

constraints may have significant impact on the mate forces (forces in

the mechanical connections, defined by the mates) and will be

discussed later in the course. The resulting force obtained for this

mechanism is, however, correct as the redundancies present in this

assembly do not have any effect on the force shown in the figure below.

Click No.

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30 Review the plot.

Play the simulation and observe the reaction force vector.

When viewed from the Right view:

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Plotting Kinematic Results

The Results PropertyManager provides access to various output

quantities reviewed in Lesson 1. They can be requested in absolute or

relative values, and with respect to another component of the assembly.

While in most situations the default output is in the global coordinate

system of the top level assembly, it is very easy to transform the values

to any other selected local coordinate system.

Absolute vs. Relative values

To request the plot of absolute values, select the

component (mate, motor, part etc.) in the


To plot values relative to the second

component, add this component to the


Note The reference component must be selected as the second in the list.

Output coordinate system

Typically, the results are output in the global coordinate system of the

assembly. For some simulation components (mates and motors, for

example) the default output is, however, in the local system of the

selected component.

To plot results in other than the default coordinate

system, select the desired component in the

Component to define XYZ directions field. The

values will then be transformed into the coordinate

system of the selected part.

Note The requested output coordinate system is indicated

by the triad shown in the graphics area.

Reference component

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In the next section four plots we will demonstrate the results in their

absolute and relative magnitudes, evaluated in both the global and local

coordinate systems.

31 Absolute result for component in global system.

Create a plot for the X component of the linear displacement of arm.

Define the plot by Displacement/Velocity/Acceleration, Linear

Displacement and X Component.

Select any face of the arm component for the Simulation element.

Click OK.

Note If we select a face, the plot will be of the linear displacement of the

part’s origin, indicated by a small blue sphere, with respect to the origin

of the assembly in the global assembly system.

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32 Absolute result for component transformed in local system.

Create a plot for the X component of the linear displacement of arm

transformed in its own local coordinate system.

Edit the definition of the previous plot and select arm as the

Component to define XYZ directions.

Click OK.

Note Note that the triad on the part arm now indicates the output local

coordinate system which is misaligned with the global coordinate

system. Further more, note that this local output system translates and

rotates with respect to the global coordinate system as you play the

motion.

Note The above figure shows the linear displacement of the part’s origin,

with respect to the origin of the assembly, transformed in the part’s

coordinate system. Alternatively, we can view the above graph as the

values from step 31, transformed in the coordinate system of the arm.

Local coordinate systemof arm part

Global coordinate system

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33 Relative result for component in global system.

Create a plot for the X component of the linear displacement of the

arm relative to the displacement of the collar.

Edit the previous plot.

Clear the Component to define XYZ directions field.

Select the collar part as the second component in the Simulation

element field.

We can see that the displacement has somewhat different oscillatory

characteristic as the displacement of the arm in the global coordinate

system (step 31). Relative result for component transformed in local

system.

Note The above figure shows the linear displacement of the arm’s origin,

with respect to the origin of the collar part in the global coordinate

system.

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34 Relative result for component in local system.

Create a plot for the X component of the linear displacement of the

arm relative to the displacement of the collar. Transform the results in

the local coordinate system of Link1.

Edit the definition of the previous plot and select Link1 as the

Component to define XYZ directions.

Note Note that the triad on the part Link1 now indicates the output local

coordinate system which is misaligned with the global coordinate

system.

The above plot shows the values plotted in step 33, transformed in the

coordinate system of the collar.

Local coordinate systemof Link1 part

Global coordinate system

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Angular Displacement Plots

Angular displacement plots can be created to measure the angular

displacement of a motor, mate, three points or one component relative

to another component. Because the angular displacement is not a

vector, only the magnitude can be plotted.

The previous section introduced generation of kinematic results plots

for a component. In the next steps various post-processing plots for

other simulation elements (mates, motors etc.) will be generated. For

most of these simulation elements, the default output coordinate system

is the local coordinate system of the element.

35 Angular displacement of mate.

Create a plot for the angular displacement of the local hinge mate

between the part Link1 and the cardian.

Define the plot by Displacement/Velocity/Acceleration, Angular

Displacement and Magnitude.

Select the local hinge mate between the Link1 and cardian for the

Simulation element.

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Click OK.

Notice the triad at the location of the hinge. It indicates that the output

coordinate system is the local system of the hinge mate. Only the

magnitude can be requested.

This plot shows the vertical rotation of the Link2 part which is

approximately 2 degrees.

36 Angular displacement of motor.

To explore the other options of the angular displacement plot, we will

modify our existing plot rather than create a new plot.

In the Results folder, right-click the last plot and click Edit Feature.

Delete the hinge mate and select the motion component Motor - crank for the Simulation element.

Click OK.

The plot shows harmonic motion of the motor. The angular

displacement goes from zero to +180 degrees then returns from -180

degrees. The slope of the graph is constant.

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37 Angular displacement of two lines defined by

three points.

This time, we will create a plot to show the angular

displacement between two lines defined by three

points.

Create a new plot.

Define the plot by Displacement/Velocity/

Acceleration, Angular Displacement and

Magnitude.

For the Simulation element, first select the two

vertices shown, then select the edge.

Select Show vector in graphics window. This

will show lines between the three selected points.

Click OK.

Vertex 2

Vertex 1

Edge(defining Vertex 3)

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The plot shows the angle between the line defined by Vertex1 and

Vertex3 and the line defined by Vertex2 and Vertex3 (Vertex3

therefore defines the center point).

Notice that in the present case the limits of the angular motion is 84

degrees and 121 degrees giving the range of 37 degrees.

Angular Velocity and Acceleration Plots

Similarly to the angular displacement, Angular velocity plot can be

generated for a motor, mate and a component relative to another

component. Magnitude as well as all three coordinate components are

available.

39 Plot angular velocity and acceleration.

Generate a couple of velocity and acceleration plots on your own. Try

to plot both the absolute and relative magnitudes in both the global and

local coordinate systems.


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Summary In this lesson we analyzed a 3dcrankslider assembly. While both

SolidWorks and SolidWorks Motion Simulation assemblies can be built

in many ways with various mates, the main objective of this lesson was

to show the suggested assembly building procedure for the motion

analysis where only kinematic results (displacements, velocities,

accelerations etc.) are of interest. We call this type of the analysis

“kinematic analysis”. “Dynamic analysis” is then a simulation where

mate forces and their distribution throughout the assembly is required.

These later types of analyses can be more intricate since the

redundancies need to be understood and addressed (redundancies are

subject of the later lessons in this course).

It was suggested that the most suitable approach to obtain kinematic

results while investing reasonable about of time in the motion assembly

building is to model the mates as closely to the real mechanical

connections as possible, i.e. all real mechanical hinges will be modeled

as hinge mates. This lesson also introduced the most common mate

types and the subject of local mates. Local mates are designed within

the SolidWorks Motion Simulation tab and do not affect the original

SolidWorks assembly and the design intent in any way. This way, each

Motion Simulation study may feature its own independent mates.

While motors and forces input may be defined in many ways, this

lesson shows the procedure to control the magnitudes using the

imported data from the table. The second half of this lesson introduces

various available result quantities and shows their definitions in detail.

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SolidWorks 2011 Exercise 2Piston

65

Exercise 2:Piston

In this exercise, we will manually create local mates and run a motion

simulation on a simple engine under the effects of gravity only. We will

plot the results and check the assembly for interference.


� Creating Local Mates on page 34.

� Angular Displacement Plots on page 59.


Open Piston from the Lesson02\Exercises folder.

2 Type of Study.

Select the Motion Study 1 tab and set the Type of Study to Motion

Analysis.


Verify that the document units are MMGS (millimeter, gram, second).

engineblock

crankshaft

conrod

piston

bearing

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Exercise 2 SolidWorks 2011Piston

66

4 Verify fixed and floating states of components.

Examine the assembly. Three parts are fixed and three parts are not

fixed and not mated. MateGroup1 is empty.

The engineblock and the two bearings are fixed.

The piston, crankshaft and conrod are floating.

5 Move components.

Move the floating components away

from their final positions. We are doing

this just to make it easier to select faces

as we create local mates.

6 Add local mates.

Add the following local mates:

� Hinge between the crankshaft and

bearing<2>.

Note The second hinge mate between the crankshaft and the bearing<1>

components could have been defined as well. However, it would have

no effect on the kinematic results of this simulation.

Fixed Components Floating Components

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� Hinge between the conrod and

crankshaft.

� Concentric between the piston

and the cylindrical face of the

engineblock (piston bore).

� Concentric between the upper

hole in the conrod and one of the wrist pin holes

in the piston. We do not have the wrist pin

modeled so we are using the concentric mate in its

place.

7 Add gravity.

Under Gravity Parameters, Direction Reference, select the Y

direction.

Under Numeric gravity value, type in a value of 9806.65 mm/sec^2.

8 Motion Study properties.

Set the study properties to record 100 frames per second.

9 Run the simulation for 2.32 seconds.

Make sure that the study type is set to Motion Analysis.

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10 Examine the motion.

Playback the study at one-quarter speed.

The weight of the piston and conrod will cause the piston to try to move

to bottom dead center. As there is no friction, the model will just

oscillate as the total energy of the system is conserved.

While the assembly moves freely, we cannot tell if there is interference

between the different components. In Lesson 3, interference detection

in SolidWorks Motion will be demonstrated.

11 Plot results.

Create a plot of the angular displacement of the crankshaft.

Initially, the plot may look odd, however if you examine it closely you

can see that the component is just rocking back and forth.

12 Angular displacement of hinge mate.

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The plot should look the same as the previous plot for the crankshaft, except that the values are of opposite sign and the graph begins at 0

degrees. This is because the displacement plot for the mate, motors and

spring features are plotted at the local coordinate system by default.

13 Plot linear displacement.

Create a plot for the linear displacement of the piston in the global

coordinate system. Plot the Y-component as this is the direction along

the axis of the piston bore.

The plot shows normal harmonic motion.

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14 Transform linear displacement plot.

Transform the above linear displacement plot into the local coordinate

system of the crankshaft.

As the local coordinate system of the crankshaft rotates, the values in

the plot are changing from positive to negative.


Summary In this exercise you analyzed a small piston assembly. The main

objective was to practice the assembly building procedure when

kinematic results are of interest only and to plot various result

quantities.

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SolidWorks 2011 Exercise 3Trace Path

71

Exercise 3:Trace Path

In this exercise we will use motors that are driven by tabular data to

have a stylus trace a path like a pen plotter.


� Local Mates on page 40.

� Importing Data Points on page 47.

Procedure Open the existing assembly from the Exercises folder.


Open pant1 from the Lesson02\Exercises folder.

2 Set the document units.


Select MMGS (millimeter, gram, second) for the Unit system.

3 New study.

Crate a new motion study. Make sure you select Motion Analysis.

chassis

cross beam

pointer

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Exercise 3 SolidWorks 2011Trace Path

72


Existing mates allow the cross beam to

move along the rails of the chassis and the

pointer to move along the cross beam.

One mate missing is something to keep the

pointer from rotating around the cross beam.

5 Add rotary motor.

To prevent the rotation, we will use a rotary motor.

Select Axis1 in the Pointer as the Component.

Set the Motion to Distance and make the distance

0 degrees from 0 to 20 seconds.

6 Add linear motor.

The first linear motor will drive the cross beam along the chassis.

Two csv files are provided in the Exercises folder, movx.csv and

movy.csv. These files have number pairs with the first number

indicating the time and the second number representing position.

Notice that in each set of numbers, the time points are evenly spaced.

This will allow us to use the Akima interpolation type.

Add a Linear Motor.

Select the face shown in the image below.

Select Data Points to open the Function Builder window.

Import Data for the Displacement (movy.csv file).

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SolidWorks 2011 Exercise 3Trace Path

73

Note Look at the triad to see that the selected face will move in the

Y direction, so we need the movy.csv file instead of the movx.csv.

Click OK.

7 Add another motor.

Add another linear motor to move the pointer across the cross beam

using the movx.csv file. Orient the motor in the direction of the

negative X axis.

8 Run the study.

Run the study for 20 seconds.

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Exercise 3 SolidWorks 2011Trace Path

74

9 Create a path trace.

Create a new result.

Select Displacement/Velocity/Acceleration and Trace Path.

Select the vertex at the end of the Pointer.

Check the Show vector in the graphics window checkbox to see the

star shape.

Note The Trace Path plot will be discussed in more in detail in Lesson 6,

where it will be used to generate the profile of a CAM.


Summary In this exercise you analyzed a pen assembly. The main objective of

this exercise was to define local mate definitions and to import

tabulated data to control the motor magnitude.

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75

Lesson 3Introduction to Contacts,

Springs and Dampers


� Check interference of components.

� Apply contact to components.

� Specify solid bodies contact friction.

� Add a spring with damper to the assembly.

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Lesson 3 SolidWorks 2011Introduction to Contacts, Springs and Dampers

76

Contact and Friction

In this lesson we will examine the motion of a catapult as it is loaded

and throws a projectile. Some of the components in this lesson are not

connected to the others through mates or joints but are restricted based

on their contact with other components. We will place these dynamic

components into our system by defining contact conditions and also

include friction between components.

Case Study: Catapult

The crank will rotate the catapult arm, through a belt and pulley, to a

position where a projectile can be loaded. The crank motion will also

be transmitted through a gear assembly to a trigger mechanism that will

release the projectile and allow the spring to push the projectile onto

the projectile holder.

When released, the counterweight will cause the arm to rotate and

throw the projectile.

Catapult-Arm

ProjectileCounterweight

Trigger mechanism

Gear assembly

Hand crank

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SolidWorks 2011 Lesson 3Introduction to Contacts, Springs and Dampers

77

Problem Description

The crank will rotate 2.75 turns to load the catapult. The motion of the

rack will cause the trigger to release the projectile onto the projectile

holder. The mechanism will release the arm and the counterweight will

cause the arm to throw the projectile.

Determine the torque required to rotate the crank and load the catapult.

Determine the displacement and velocity and force of the loading

spring.


� Create a Motion Study.

This will be a new motion study.

� Apply friction.

Friction will be added to the existing SolidWorks mates.

� Apply contact.

Contact will be added to the dynamic components.

� Add a spring.

We don’t use a spring model in the motion simulation. Instead we

create a motion element that mathematically represents the spring.

� Apply gravity.

The catapult operates under conditions of normal gravity.

� Calculate the simulation.


We will create various plots to show the torque and power required.


Open Catapult-assembly from the Lesson03\Case Studies folder.


The crank rotation does two things,

it rotates the arm through the belt

and pulley and it triggers the release

of the projectile through a gear

train.

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78

The gear train consists of

three gear mates and a rack

and pinion mate.

When the rack moves, it

will come in contact with

the release mechanism and

lift the projectile holder

door.

Rotate the crank to see how the mates work.


Verify that the document units are set to MMGS (millimeter, gram,

second).


Right-click the Motion Study 1 tab and click Create New Motion

Study.



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5 Add a motor.

To revolve the crank we need to apply a motor to

the end of the shaft. We want to rotate the crank

2.75 turns in 3 seconds.

Click Motor in the MotionManager toolbar.

Select the edge of the crank shaft for both the

Motor Direction and Motor Location fields.

Select Rotary Motor for the Motor Type and Distance for the Motion

Type.

Type 990 deg (2.75 turns x 360 deg) for the Displacement and

3 seconds for the Duration.

Click OK.

6 Disable the motor.

After the motor turns for 3 seconds, we want it to hold the catapult in

the loaded position while the projectile moves into the projectile tray.

We then want the motor to disengage to allow the counterweight to

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The entire simulation will run for 5 seconds, so to make it easier to

select on the timeline, click Zoom In on the lower right corner of

the MotionManager until a little more than 5 seconds fills the time line

MotionManager.

Select the RotaryMotor1 in the MotionManager. Right-click in the

timeline at 3.4 seconds and click Off. This creates a key that suppresses

the motor at 3.4 seconds so that it will have no effect after this time.

Tip If you place the key at the wrong time, just drag it to 3.4 seconds.

7 Motion Study Properties.

Set the Frames per second to 50.

8 Calculate.

Click Calculate and observe the motion.

Notice that as specified, motor rotates the crank by 2.75 turns in

3 seconds. From 3 to 3.4 seconds the motor keeps the crank, as well as

the arm, stationary and ready to launch. Finally it disengages at

3.4 seconds when the mechanism begins to move in not specifically

defined motion. A few key elements must still be added to the motion

model, however.

9 Analyze the motion.

In the MotionManager, right-click Orientation and Camera Views

and click Disable Playback of View Keys.

Change to the Front view and zoom in on the left end of the assembly.

Play the simulation in slow motion again and notice that the two

triggers move through each other.

To stop this, we must add contact between them. Before defining

contact, we will however introduce a feature which can be used to

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81

Interference Detection

The Interference Detection tool in SolidWorks will detect interference

between components. However, it will only detect interference for a

single static position of the components.

In SolidWorks Motion, interference can be detected for the motion path

of each component.

Where to Find It � Right-click on the top level component in the MotionManager and

select Check Interference.

10 Check Interference.

In the MotionManager, right-click the Catapult-assembly and click

Check Interference.

Select the two triggers and click Find Now.

11 Examine the results.

The two triggers interfere starting at Frame 132 at time 2.620 seconds

and remain that way until the last frame.

Select the first interference and click Details. We

can now see the location and amount of

interference.

Close the dialog Find Interferences Over Time.

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82

Contact Contacts can be defined between multiple bodies or curves to prevent

penetration. In this lesson we will only learn how to define the contact

between solid bodies and discuss friction. A more detailed discussion

on the definition of contacts and its parameters will be presented in the

next lesson.

Introducing:

Contact

Contact is used to define the way bodies react with each other. Within

the contact definition, we can control the friction and the elastic

properties between the bodies.

Where to Find It � Click Contact on the MotionManager toolbar.

12 Add contact.

In the MotionManager toolbar click Contact .

Select the two Projectile holder trigger parts.

For Contact Type select Solid Bodies.

We will keep all contact parameters except friction

at their default values - they are the subject of

Lesson 4.

Make sure that Material PropertyManager is

checked and select Steel (Dry) for both materials.

We will run the simulation without considering the

friction between these two parts. Clear Friction.

Click OK.

13 Calculate.

14 Examine the trigger.

When the simulation runs, the trigger on the rack

mechanism will now contact the trigger on the

projectile holder door and lower it.

Contact groups Contacts between the bodies can be defined in a multiple separate

definitions (each for two bodies only), or in one (or a few) definition

with all bodies included in a single (or a few) definition only. The later

one will consider contact between all selected bodies, thus

automatically generating multiple contact pairs. While this procedure is

easy to define, considering contact for all pairs can be computationally

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Contact definition with contact groups ignores contact between parts in

the group, but considers contact between all combinations of pairs of

bodies across the groups. It is possible to define a maximum of two

contact groups.

Introducing:

Use contact groups

Contact groups enable to place contacting bodies in two separate

groups. All contact combinations across the two groups are considered

only.

Where to Find It � Click Contact on the MotionManager toolbar. Under

Selections, check the Use contact groups check box.

15 Additional contacts.

The following additional contacts have to be defined:

� projectile - projectile holder door� projectile holder - projectile� projectile holder pusher - projectile

Using the procedure outlined in the previous steps, three separate

definitions would have to be created. Instead, using the contact groups

one single definition will suffice in this case.

Create contact definition. Under Contact Type

select Solid Bodies, for Material select Steel

(Dry) and clear Friction.

Under Selections check Use contact groups.

Select Projectile in Group1 and projectile holder, projectile holder door and projectile holder pusher in Group2.

Click OK.

Note The PropertyManager indicates that three contact pairs will be considered for the calculation.

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Contact Friction When defining contact, there are three friction options which can be

used depending on the model.

� Static

� Kinematic

� None

Once you decide what friction types to include in your contact, you

must evaluate the static and/or kinematic velocity and friction

constants.

Coulomb friction forces are calculated based on two different

coefficients - static and kinematic.

Static Coefficient The static coefficient is the constant used to calculate the force

necessary to overcome friction when a body is at rest.

Kinematic

Coefficient

The kinematic coefficient is the constant used to calculate friction

forces once the body is no longer at rest.

In reality, the static friction transition velocity is zero, but numerical

solvers, such as SolidWorks Motion, require that a non-zero value be

specified to avoid singularity at the origin. More specifically, when a

part is in transition from a negative to positive velocity, and when the

velocity is zero, the force magnitude cannot instantaneously transition

from a positive to negative value.

Therefore, the graph

shows how

SolidWorks Motion

resolves this issue—

the user specifies a

static and kinematic

transition velocity

where the friction

coefficients are used.

From there,

SolidWorks Motion fits a smooth curve to solve for the friction force.

In the graph above, the default friction parameters for dry steel in

contact are used.

� Static Friction Transition Velocity: vs = 0.102 mm/s.

� Kinematic Friction Transition Velocity: vt = 10.16 mm/s.

� Static Friction Coefficient: 0.30.

� Kinematic Friction Coefficient: 0.25.

Where to Find It � In the PropertyManager for Contact, select Friction.

Velocity (mm/sec.)

10.16 mm/sec.0.102 mm/sec.

Static

Kinematic

Force (N)

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16 Additional contact set.

Create a contact set between:

� projectile� Catapult-Arm

Under Material select Steel (Dry).

Make sure that Friction is checked.

Uncheck Material and change the values for the

Dynamic Friction Coefficient and Static

Friction Coefficient to 0.15 and 0.2, respectively.

Click OK.

Note Unchecking Material opens up the Friction dialog fields for editing. The contact characteristics in Elastic Properties, determined by the selection in the Material dialog, remain unchanged. Elastic Properties are discussed in Lesson 4.

Translational Spring

A translational spring represents the displacement dependent force

acting between two parts over a distance and along a particular

direction.

When defining a spring, you can readily change the force-displacement

dependency from linear to another predefined relationship by selecting

the function type from a list. This allows you to select the relationship

between the force and displacement. The following force-displacement

relationships are supported in SolidWorks Motion:

X, X2, X3, X4, 1/x, 1/x2, 1/x3

You specify the location of the spring on two parts.

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SolidWorks Motion calculates the spring force based on the relative

displacement between the two parts, the stiffness of the spring and the

fabrication or free length.

When the spring force is negative, the spring is in a stretched position

relative to the free length.

Note Spring forces become ill-defined if the end points become coincident

because of undefined direction.

Magnitude of Spring Force

The magnitude of the spring force is based on the stiffness and initial

force.

The spring relationship can be written as:

F = -K (X - X0)n + F0

Where:

X= Distance between the two locations that define the spring

K= Spring stiffness coefficient (always > 0)

F0 = Reference force of the spring (preload)

n = Exponent. For example, if spring force = KX2, then n = 2. Valid

values for the exponent n are: -4,-3,-2,-1,1,2,3,4.

X0 = Reference length (at preload, always > 0)

� Positive force repels the two parts.

� Negative force attracts the two parts.

Note To create a spring that exhibits non-linear force properties not

supported in the spring definition, you must use an action-reaction

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Introducing: Spring Both linear and torsional springs can be added between components.

Both the Exponent of the spring force expression (linear to ± 4) and

Spring Constant can be specified.

Where to Find It � Click Spring on the MotionManager toolbar.

Translational Damper

A translational damper is considered a resistive element used to

“smoothen” out oscillations encountered due to outside forces.

Typically, dampers are used in conjunction with springs to “dampen”

out any oscillations or vibrations created by the spring.

In the real world, bodies and even springs have built in structural

damping, and the damper element can be used to represent this. The

force created by a damper is dependent on the instantaneous velocity

vectors between the two defined endpoints.

Note To create a damper that exhibits non-linear force properties not

supported in the Damper definition, you must use an action-reaction

force where you can enter a non-linear force equation based on the

velocity between the two points of the force entity.

For the translational damper element, the force equation is pre-defined

as where c is the user defined damping coefficient, v is the

relative velocity between two end points and n is the exponent. For

example, if damper force = -c*v2, then n = 2 (valid options are -4,-3,-

2,-1,1,2,3,4).

Introducing: Damper The Damper can be added between components in a mechanism.

Additionally, both linear and torsional springs can have damping

properties that act as the combination of spring and damper together.

Like springs, both the Exponent of the damper force expression

(linear to ± 4) and Damping Constant can be specified.

Where to Find It � Click Damper on the MotionManager toolbar.

F c vn

×=

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17 Add a spring.

To move the projectile into position on the

Catapult-Arm, we must add a spring. The spring

will have a pre-load to hold the projectile against

the back of the projectile holder door. When

the door drops, the projectile is pushed into

position.

Click Spring on the MotionManager toolbar.

Select the two faces shown below.

Set the Spring Parameters as shown to create a linear spring with a

Spring Constant of 0.15 N/mm and the Free Length of 13mm.

Select Damper and add a Damping Constant of 0.01 N/(mm/s).

For Display set the Coil Diameter to 4.00mm, 5 turns and a Wire

Diameter of 0.5mm.

Note The values entered in the Display area are only used as graphics

parameters.

Click OK.

18 Calculate.

When the simulation solves, the projectile flies off into space and arm

does not release and the counterweight does not stay level. This is

because we are still missing a key element, gravity.

19 Add gravity.

Add gravity to the assembly.

20 Calculate.

This time the arm is cranked down to the loading position and is held

there by the motor while the trigger releases the door and the projectile

is pushed onto the arm by the spring. At 3.4 seconds, the motor turns

off and the gravity on the counterweight swings the arm and launches

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Post-processing

Now that the simulation is calculated, we can create plots for the

different parameters we are interested in.

21 Motor torque.

Create a new plot.

Define the plot using Forces, Motor Torque and Magnitude.

Select the RotaryMotor as the rotational element.

We observe that the top torque magnitude reaches approximate 7 N-

mm.

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90

22 Spring displacement.

Create a new plot.

Define the plot using Displacement/Velocity/Acceleration, Linear

Displacement and Magnitude.

Select the Linear Spring as the simulation element.

The spring expands from 6 to 13 mm. In the setup of the problem, we

specified the length of the uncompressed spring as 13 mm.

23 Spring velocity.

Create a new plot.

Define the plot using Displacement/Velocity/Acceleration, Linear

Velocity and Magnitude.


From the plot, we can see that the spring reaches a top speed of 91 mm/

sec.

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24 Spring force.

Create a new plot.

Define the plot using Forces, Reaction Force and Magnitude.


Click OK.

25 Warning.

Similarly to Lesson 2, we will receive a warning about redundant

constraints. Redundant constraints may have significant impact on the

mate forces (forces in the mechanical connections, mates, defined by

the mates) and will be discussed later in the course. The resulting force

obtained for this mechanism is, however, correct.

Click No.

26 Review the plot.

From the plot we can see that the maximum spring force is 1 N.

We can see that the spring only pushes the projectile for about

0.1 seconds.

Analysis with Friction (Optional)

In this part we will study the effect of contact friction on the motion of

the projectile. We will use the study we have just done and add friction

between the projectile and the projectile holder.

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27 Duplicate the study.

Duplicate the existing motion study and name it Larger Friction.

28 Add friction.

Edit the group contact set containing the projectile and projectile holder. Activate Friction with the default values for Steel (Dry).


Under Motion study Properties set Number of Frames to 100.

Click the Advanced Options button and change the Integrator Type

to WSTIFF.

Note Integrators are discussed in detail in Lesson 4.


31 Animate results.

Animate the results and

notice that the projectile

would not slide onto the arm

due to the added friction.


Summary In this lesson we analyzed a catapult assembly. The main objective was

to rotate the arm to the position where a projectile can be loaded, then

release the arm and eject the projectile. The following features were

used and explained in detail: interference check through the computed

time steps, definition of the spring and damper and the specification of

the solid body contact with the contact groups. Because the parameters

of the contact setup are subject of Lesson 4, this lesson only introduced

the procedure to define the contacts with friction. Both static and

kinematic friction types were introduced and shown. This assembly

also features multiple gear mates.

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SolidWorks 2011 Exercise 4The Bug

93

Exercise 4:The Bug

In this exercise, we will use a mechanical bug with an oscillating motor

to demonstrate the effects of friction on the movement of parts. We will

run the study twice, first without friction and then with friction.


� Contact Friction on page 84.


Open Bug Assembly file from the Lesson03\Exercises folder.

The assembly consists of a flat plate and a two piece mechanical bug.

The intent is to have the movement of the leg move the bug along the

plate. There is a Coincident mate between central planes on the Base

and Plate to keep the Bug moving down the middle of the Plate.



Verify that MMGS (millimeter, gram, second) is selected for the Unit

system.

3 New study.


4 Add gravity.


5 Add contact.

Using contact groups, add solid body contact between the Plane and

the two parts of the bug (Leg and Base).

Select Rubber (Dry) for the material.

Clear Friction.PRE-

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Exercise 4 SolidWorks 2011The Bug

94

6 Add a motor.

Add an oscillating Rotary Motor to

the Leg. Attach the motor to the

edge shown and set the motor to

move 30 degrees at 5 Hz.

7 Calculate.

Calculate the analysis for

2 seconds.

While the motor oscillates properly,

without friction, the bug does not

move.

8 Add friction.

Edit the two contacts and select Friction. The dynamic friction

coefficient will be set to that of the specified material (Rubber (Dry)).

Select static friction and use the default values.

9 Re-calculate.

Run the analysis for 20 seconds.

With friction added, the bug will move along the plate.


Summary In this exercise you analyzed a small assembly called bug. The main

objective was to see the effect of the friction model in the contact

specification. While in the model without friction the bug assembly

does not move, addition of the friction come close to reality – the bugs

moves along the base plane.

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SolidWorks 2011 Exercise 5Door Closer

95

Exercise 5:Door Closer

In public buildings such as schools

or offices, door closers are often

added to non-motorized swing

doors to ensure that the doors

automatically close after use. To

ensure that the doors do not close

too quickly and slam, a spring

damper is added to the interior of

the door closer.

Door Closer Analysis

In this exercise, we will use the Motion Manager to add an internal

spring and damper to the door closer. We will then use SolidWorks

Motion to plot the effect of the spring and damper on the door's

behavior and adjust the parameters to achieve the desired result.


� Translational Spring on page 85.

� Translational Damper on page 87.

Procedure Open the existing assembly from the Exercises folder.


Open door from the Lesson03\Exercises folder.



Verify that MMGS (millimeter, gram, second) is selected for the Unit

system.

3 New study.


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Exercise 5 SolidWorks 2011Door Closer

96

4 Create a linear spring.

Define a Linear Spring between the gas-piston and gas-cylinder.

Use the circular edges as indicated in the figure. You must select the

edges and not the faces or else the software does not use the center. The

spring must be aligned with the cylinder.

Use 1 N/mm and 180 mm for the Spring Constant and the Free

Length, respectively.

Use 5 N/(mm/s) for the Damping constant. Input appropriate values

in the Display PropertyManager.

Note It may be necessary to change the transparency of the door closer's

gas cylinder in order to select the interior parts necessary to define

the linear spring.

Click OK.

Note The damper is used to prevent doors from slamming shut due to the

force of the spring.

5 Run the Motion Analysis.

Run the analysis for 40 seconds.

Edges for the spring definition

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SolidWorks 2011 Exercise 5Door Closer

97

6 Plot door velocity.

Create a plot of the door (center of mass) velocity magnitude.

Notice that the door closes too quickly (within approximately 24

seconds) and passes through the door frame before coming to a

complete stop.

We do not wish to close the door so quickly. Furthermore, we do not

want the door to actually pass through the door frame and open on the

opposite side. To solve this, we need to redefine the spring and damper

constants.


Note It is possible to simply change the constants in the Motion Study we

just created. However, we want to be able to compare results from the

two constant settings. Therefore, we will duplicate the initial Motion

Study and make modifications to the duplicate study.

8 Redefine the spring with damper.

Increase the Spring Constant value from 1.00 N/mm to 2.00 N/mm.

Increase the Damping Constant value from 5.00 N/(mm/s) to

10.00 N/(mm/s).

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Exercise 5 SolidWorks 2011Door Closer

98

9 Calculate the Motion Analysis.

Calculate and plot the door velocity.

10 Compare results.

Clicking either of the motion studies we just completed will enable you

to compare the results from both studies. You can observe that in the

second study, the door closes slower and comes to a complete stop

without actually passing through the frame.

Conclusion From the data in the two simulations, we can determine the appropriate

spring and damper constants for the door to close as desired and

without slamming.

Summary In this exercise you analyzed a door assembly. The main objective was

to practice the definition of the spring and damper to model the door

closer and to find an optimum spring and damper parameters to close

the door slowly without it passing through the frame.

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99

Lesson 4Advanced Contact


� Understand the definition as well as the description of contacts.

� Use expressions to prescribe the magnitude of forces and motors.

� Analyze some causes of the incorrect solution or a contact solution

failure.

� Use alternative numerical integrators.

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Lesson 4 SolidWorks 2011Advanced Contact

100

Contact Forces The objective of this lesson is to get familiar with the definition of solid

body contacts, as well as understanding their limitations and use in

SolidWorks Motion. The expressions utilizing various mathematical

functions prescribing displacements and other study features will be

introduced. Contact force as the latch closes and the force needed to

close the latch will be extracted; accuracy of the contact force will be

discussed as well.

Case Study: Latching Assembly

In this assembly, an

over-center latch is

used to hold the

Carriage part against

a spring.

Problem Description

For the latching mechanism, determine:

� The contact force generated on the Spring Lever and Keeper as

the latch closes.

� The forced needed to close the latch.


Open Full Latch Mechanism. from Lesson04\Case Studies

folder.


The assembly has several mates however not all components have

enough mates to allow the parts to move based on the mechanical

motion of the final assembly.

The Carriage part is concentric to the center spindle, but can rotate

through the side spindles.

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SolidWorks 2011 Lesson 4Advanced Contact

101

Three components of the latch, knurled_pin, spring and Series Lever are not restrained laterally.

3 Verify the units.

Verify that the document units are set to MMGS.

4 Create a new Motion study.

Name the study Tessellated geometry and set Type of Study to

Motion Analysis.

5 Center the latch.

Add a Coincident mate

between the Front planes of

the Base and Series Lever.

This is a local mate. If you select the Model tab in the MotionManager,

the Series Lever can still move.

We could add another mate to restrict the motion of the J_Spring. In

the next few steps we will practice an alternative approach to constrain

the motion of free parts.

Fixing Motion with Motors

An alternative approach to additional mates is the addition of a motor.

The advantage of such an approach may not be immediately apparent,

but we will use it in this motion model.

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One reason for using a motor instead of a mate is that it does not

introduce additional constraints to the motion model and helps to

reduce the number of the redundant constraints. Redundant constraints

will be discussed in Lesson 8: Redundancies.

6 Restrict the linear translation of the latch.

Create a Linear Motor.

Attach the motor to the face shown.

For Motion, select Distance and set it to 0 mm.

Set the Start Time to 0s and the Duration to 3.5s.

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7 Restrict rotation of the Carriage.

Create a Rotary Motor.

Attach the motor to the edge shown.

For Motion, select Distance and set it to 0 deg.

Set the Start Time to 0s and the Duration to 3.5s. The simulation will

run for 3.5 seconds, so this motor will stop the Carriage from rotating

during the entire simulation.

Motor Input and Force Input Types

SolidWorks Motion allows you to set the motor input to a number of

different types. We have used Constant Speed, Distance and Data

Points in most of our lessons thus far, but Expression, Oscillating

and Segments are also available.

Expression lets us to define a profile that dictates the motion of the

motor with a help of various mathematical functions.

Functional Expressions

You can use functional expressions to define magnitudes of input used

in:

� Motors

� Forces

Functions can depend on time or other system states, such as

displacement, velocity, and reaction forces and may be composed of

any valid combination of simple constants, operators, parameters, and

available supported solver functions such as Step (STEP) and

Harmonic (SHF), for example. For a detailed list of functions and its

syntax, please refer to the on-line help.

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The following is a list of accepted functions:

Function Definition

ABS Absolute value of (a)

ACOS Arc cosine of (a)

AINT Nearest integer whose magnitude is not larger than (a)

ANINT Nearest whole number to (a)

ASIN Arc sine of (a)

ATAN Arc tangent of (a)

ATAN2 Arc tangent of (a1, a2)

COS Cosine of (a)

COSH Hyperbolic cosine of (a)

DIM Positive difference of a1 and a2

EXP e raised to the power of (a)

LOG Natural logarithm of (a)

LOG10 Log to base 10 of (a)

MAX Maximum of a1 and a2

MIN Minimum of a1 and a2

MOD Remainder when a1 is divided by a2

SIGN Transfer sign of a2 to magnitude of a1

SIN Sine of (a)

SINH Hyperbolic sine of (a)

SQRT Square root of a1

STEP Smoothed step function

TAN Tangent of (a)

TANH Hyperbolic tangent of (a)

DTOR Degrees to radians conversion factor

PI Ratio of circumference to diameter of a circle

RTOD Radians to degrees conversion factor

TIME Current simulation time

IF Defines a function expression

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Force Functions There are five types of force functions that can be used to define the

force:

� Constant: Sets a constant value.

� Step: Defines a step by an Initial Value, Start Time, Final Value,

Final Time.

� Harmonic: Defines the value by Amplitude, Frequency, Average

and Phase Shift.

� Segments: Defines the value by combining segments of the most

commonly used functions such as linear, polynomial, half-sine and

others.

� Data Points: Takes the values from a table of data points and

interpolates a spline between the data points.

� Expression: Defines the value using a formula.

STEP Function A STEP function prescribes the given quantity (displacement, velocity,

acceleration or force magnitude, for example) between two values with

a smooth transition. Before and after the transition, the displacement,

velocity or acceleration magnitude is constant.

For example, consider the

illustration at the right where:

d0 = Initial value of displacement

d1 = Final value of displacement

t0 = Start step time

t1 = Final step time

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8 Create a rotary motor to drive the latch.

Hide the J_Spring.

In the Motion Manager, click Motor .


Under either the Motor Location or Components/Direction fields,

select Axis1 of the Series Lever as indicated in the figure. This motor

will simulate the action of the hand operating the Series Lever to open

and close the latch.

Under Motor Type, in the Motion field, select Expression. The

command brings up the Function Builder window.

9 Build motor expression.

In the Function Builder, make sure that the Expression button is

selected.

Select Mathematical Functions for the input type and double-click

STEP(x,x0,h0,x1,h1) to insert the step function.

Modify the functional expression to read STEP(TIME,0,0D,1,90D).

Note The TIME variable can be typed in or inserted by chancing the input

type to Variables and Constants and double-clicking TIME.

Complete the expression to its final form:

STEP(TIME,0,0D,1,90D)+STEP(TIME,1.5,0D,3,-90D)PRE-

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Note The Function Builder graph windows will update the plots for

displacement, velocity, acceleration and jerk automatically.

Click OK to complete the definition of the expression and close the

Function Builder.

Click OK to complete the definition of the Motor feature.

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Note The above expression is a

combination of two step functions.

The first rotates the Series Lever

component by 90 degrees between 0

and 1 second and then it keeps the

vertical position for 0.5 seconds until

the time 1.5 seconds.

At time 1.5 seconds, we add the

second step function which changes

the rotational displacement back to

zero between the 1.5 and 3 seconds.

Both functions as well as the

combination (the final motion of the

Series Lever) are shown in the

figures.

10 Define Spring and

Damper.

We now need to define a spring with a damper which generates tension to keep the latch pulled tight.

In the Motion Manager,

click Spring .

Choose a Linear

Spring with a spring constant of 10 N/mm, and create the spring at the locations shown in the figure below.

Keep the Free Length at its default value.

Turn on the linear Damper and specify a magnitude of 0.10 N/(mm/s).

Notice that the free length of the spring is automatically populated into the Free Length field.

Click OK.

STEP(TIME,0,0D,1,90D)

STEP(TIME,1.5,0D,3,-90D)

Combined

1.0 sec

1.5 sec

3.0 sec

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109

Contact: Solid Bodies

Contacts are defined between two or more bodies or two curves (a

contact pair). During the definition of the contact between solid bodies,

whatever feature you pick on the parts, the corresponding body will be

selected (and used for the contact analysis). During the solve, the

software calculates at each frame the bounding boxes of the parts

interfere. As soon as it is the case, a finer interference calculation is

done between the two bodies and from the center of gravity of the

interference volume, an impact force is computed and applied on both

bodies.

This procedure is schematically shown in the figure below.

To understand the contact treatment in the SolidWorks Motion, we first

need to reiterate the very original assumption of this modulus: all parts

participating in the motion simulation are rigid. Contact conditions are

used to simulate impact of the two or more colliding parts (which are

not rigid in real life). Nearly without exceptions all impacts feature

high relative velocity, which result in elasto-plastic deformations with

severe localized strains and significant changes in the local geometry

(geometry of the contact region). Approximations are therefore

necessary.

SolidWorks Motion allows for the specification of the contact

parameters using two distinct approaches: Impact properties (Impact

force model) and Restitution coefficient (Poisson model).

1. 2. 3. 4.

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Poisson Model (Restitution Coefficient)

Restitution coefficient (Poisson model): Poisson model is based on the

utilization of the restitution coefficient e is defined in the following

relationship:

Where v1 and v2 are the

velocities of the spheres

before the impact and v1’ and

v2’ are the velocities after the

impact. The bounding values

of this coefficient are (0;1),

where 1 indicates perfectly

elastic impact where no

energy is lost, while 0 indicates perfectly plastic impact where the parts

adhere after the impact and maximum possible energy is lost.

The restitution coefficient is geometry dependent and spheres in the

above illustration are used for the demonstration purposes only.

Poisson model does not require specification of the damping coefficient

(as is the case of the Impact force model, discussed below) and does

account correctly for the energy dissipation. The use of this model is

therefore recommended if energy dissipation is of the great importance

in the simulation. Also, determination of the Poisson model parameters,

restitution coefficient e, is more straight forward than in the case of the

Impact force model; in many instances, the restitution coefficient can

be measured using the standardized methods (see ASTM F1887-98

Standard Test Method for Measuring the Coefficient of Restitution

(COR) of Baseballs and Softballs, for example) or found in various

tables. This model is not suitable for the persistent impacts (impacts,

where contact is developed for a prolonged periods of time); Impact

force model should be used instead in these situations.

Impact Force Model

Impact properties (Impact force model): Impact properties in

SolidWorks Simulation allow for the calculation of the contact force

using the following expression:

where k represent the stiffness of the contact, e is the elastic force

exponent, and c is the damping coefficient (cmax) is then the maximum

possible damping coefficient). As in the case of the restitution

coefficient, these parameters are both material and geometry dependent

and can not be apparently found in the material tables. The following

sections describe the Impact force model parameters in more detail.

v′2 v′1– e v1 v2–( )=

v1

v1’

v2’

v2’

Fcontact k x0 x–( )e

c v•–=

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Stiffness k To correctly determine the stiffness, possible solution is to model the

configuration of the contact in SolidWorks Simulation finite element

software, apply any force in the direction of the impact and solve for

the displacements. Stiffness can then be readily obtained from the force

magnitude and the resulting displacements. A figure below

demonstrates the impact configuration of two spheres meshed in

SolidWorks Simulation software.

In many instances, the elastic solution can be found in various

engineering publications. It is apparent that the computation of the

contact stiffness k can be a daunting task and simplifications have to be

introduced.

Exponent e This parameter controls the degree on nonlinearity in the elastic force;

e=1 then constitutes a linear elastic force.

Damping

Coefficient c and

Penetration d

When two objects collide and deform, portion of the kinetic energy is

consumed on the plastic deformation, heat and similar phenomena.

Approximately, this value can be obtained from the results of the

nonlinear dynamic solution (of the above problem of the two spheres,

for example) with advanced material models. Utilizing this procedure

is, however, unrealistic and simplifications are necessary. It is assumed

that the damping coefficient (a measure of the capacity to dissipate

energy) increases from zero (at the beginning of the impact) to its

maximum value cmax, when certain specified deformation is achieved;

we call this deformation value penetration d. For any deformation

larger than the penetration d, the damping coefficient is constant and

equal to cmax. A typical value for the maximum damping coefficient

cmax is 0.1% - 1% of the contact stiffness k.

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Closing Remarks It is now apparent that the determination of the above parameters is

non-trivial, time consuming and significant simplifications have to be

introduced. A corollary of the foregoing is that the solution of the

collision characteristics (impact forces, accelerations of the impacting

regions and etc.) can only be approximate. Their accurate magnitudes

can only be determined by more advanced computational methods,

such as nonlinear dynamic solutions using SolidWorks Simulation

Premium package, which can be computationally very demanding.

It is important to clarify that for the purpose of this section, impact

force and the acceleration of the impacting regions terms represent the

contact quantities at the onset of the contact where severe deceleration

forces are encountered, i.e. impact or collision. The duration of these

collisions is typically very short. After a certain time, when the

impacting or colliding components are touching and the dynamics

aspects of the solution is less important, contact forces are accurate and

can be extracted from SolidWorks Motion. This is demonstrated at the

end of this lesson.

In conclusion, if an important objective of the motion simulation is to

obtain the impact quantities (impact force, impact region acceleration

etc.), time needs to be invested in the determination of the above

parameters, or more advanced analysis type must be carried out.

Typically, users are not interested in the accurate impact region results

but rather they need to determine the kinematics or dynamics of large

systems. Approximate values are then used for the contact

characteristics and accurate solution of the system kinematics and

dynamics can be carried out efficiently.

To assist users with the impact characteristics, SolidWorks Motion

contact library features approximate values for some contact material

configurations (note that the geometry is not clearly defined). You may

use these values as a base line if the material composition of your parts

participating in the contact is similar. However, if more accurate impact

solution is needed, correct impact parameters have to be determined.

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11 Define contact between latch and latch keeper.

In the Motion Manager, click Contact .

Under Contact Type select Solid Bodies.

Select the latch arm (J_Spring), the latch lever (Lever), and the latch

keeper (keeper).

Select Specify Material to allow us to define the impact parameters.

Select Steel (Dry) from the list for both materials. Keep the Friction

on at its default values.

Here we are trying to make the impact more realistic by simulating two

hard metals colliding. As discussed above, the elastic properties of the

contact are only approximate. More realistic values would be required

for a contact region solution (contact reliable force and acceleration of

the contact region).

Click OK.

12 Define gravity.

Define gravity in the negative X direction.

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13 Motion analysis Properties.

Verify that Frames per second is set at the

default value of 25.

In the Motion Study Properties, set the 3D Contact

Resolution slider all the way to the left, to its

lowest resolution setting.

Note The contact resolution parameters are explained in the discussion

below.


Notice that the solution was achieved, but

is incorrect. The Spring passes through the

other components without developing any

of the specified contacts. There can be a

few reasons for such behavior:

� The time step of the integrator (solver) is too large, in which case

the contact is not even detected.

� The accuracy setting is too high or too low.

� The geometrical description of contact is insufficient.

In the present case it is the last one causing the incorrect solution.

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Geometrical Description of Contacts

SolidWorks Simulation treats the geometries of the contacting solid

bodies in two distinct ways:

� Tessellated geometry

The surfaces of the contacting bodies are meshed with the triangular

elements to simplify the description. The density of the mesh, i.e. the

contact geometry resolution, is controlled with the 3D Contact

Resolution parameter in the study properties. Because this description

is very efficient, yet typically sufficient to obtain accurate solutions,

tessellated geometry is the default choice. Very coarse description may

result in inaccurate solution or even failing to develop the contacts.

This is also the cause of the solution failure in the present case.

� Precise geometry

If the tessellated geometry description if not sufficient (solution is not

sufficient or can not be obtained), Use Precise Contact option can be

used instead. Exact description of the bodies’ surfaces is then used.

While this is the most accurate description, it can be computationally

expensive and should be used with caution. Use this option if your

contacting bodies feature complex or point like geometries.

Examples of the tessellated geometries at two resolution levels as well

as the precise geometry are shown below.

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15 Adjust the Study Properties.

Improve the accuracy of the tessellated data.

In the MotionManager click Motion Study

Properties .

Move the 3D Contact Resolution slider to the

right to a value of 94.

Click OK.


Notice that the computation is noticeably slower.

The simulation will fail and display the following message:

The solver failed to converge. Possible causes are:

1. The solver is failing to achieve the specified accuracy. Relax the Accuracy setting in Motion Analysis Properties.

2. If parts in the model are moving quickly, evaluate the Jacobian more often.

3. The mechanism may be getting locked. Start the simulation with a different initial configuration or change you motors to get valid motion.

4. If the failure is happening right at the beginning of the simulation, use a smaller Initial Integrator Step Size.

5. Try to use a stiff solver like ‘WSTIFF’.6. Try to avoid sharp discontinuities in the model like

sudden motion changes, force changes or mate activation/deactivation.

7. You may have motors with very high speeds. Try to reduce the motor speed.

8. Make sure that only one motor is driving a given component at any time.

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After reviewing the message, there are some possible problems. The

first possible problem is item 1, where the solver is failing to achieve

the specified accuracy. We will try decreasing the accuracy of the

solution.

The second message suggests that if the parts move too quickly, the

Jacobian should be evaluated more often. Since the Jacobian setting is

already at its maximum value, we will achieve this by also reducing the

Maximum Integrator Step Size in the Advanced Options of the

Motion Study Properties.

The point where the solution fails is when the latch reaches the over

center point because of instability in the solution.

17 Adjust the study properties.

We will reduce the accuracy in order to let the

solver handle the over center solution.

In the MotionManager click Motion Study

Properties .

Reduce the Accuracy to 0.001.

Set the Frames per second to 120 to save more

instances of data on the disk.

Click Advanced Options and reduce the

Maximum Integrator Step Size to 0.001.

Click OK to close the Advanced Motion Analysis Options.

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Accuracy and the Maximum Integrator Step Size parameters have

significant impact on the contact solution and should be used first when

contact solution problems occur. To read more about the advanced

motion analysis options, integrators and their options, refer to Appendix

A: Motion Study Convergence Solutions and Advanced Options.


This time the simulation will run, but it may take several minutes to

complete.

19 Animate.

Play the animation and zoom in on the latch mechanism.

Notice that when the latch is closed, there is a small oscillation because

all the energy is not being damped. This does not happen in the

physical model and is a sign that the damping values used in this

simulation can be increased to represent the real situation more closely.

Instability Points

Instability points can be defined as instances where self equilibrated

structure does not move, however a small impulse in either direction

will result in rapid motion during which the stored elastic energy is

rapidly transformed in kinetic energy. Such instances are difficult to

overcome numerically. This point is featured in our solution and the

solver is expectedly facing difficulties. Also, notice the time required

for the solution to complete.

20 Plot contact forces.

Plot the contact forces between the

latch and the keeper.

Create a new plot.

Define the plot using Forces,

Contact Force and Magnitude.

Select the two faces shown.

Tip To make it easier to select the faces, move the timeline to a position

where the components are in the position shown.

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We can observe that the graph exhibits significant oscillations with

apparently unbounded peaks. This interval (approximately 2.85 to 3.5

seconds) corresponds to the small oscillations observed when the latch

mechanism is closed, discussed in Step 19. Each one of those peaks

correspond to an impact (or collision) force, magnitude of which

depends nearly exclusively on the contact stiffness characteristics.

Because these are highly approximate, the peaks of the impacts forces

in this interval should be ignored.

Modifying Result Plots

Default plots are created with the X axis showing the duration of the

simulation and the Y axis scaled to the maximum value of the variable

being plotted. There are times when we want to scale the plots

differently.

Introducing: Chart

Properties

Almost all aspects of a plot can be modified, from the titles, to

background color to axis values and titles.

Where to Find It � Right-click on the time in the plot and select Chart Properties.

22 Modify the plot.

Right-click on the X axis of the plot and click Axis Properties. Select

the Scale tab.

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Clear End Point and type 3 for the new value of the end point of the

X axis.

Use the same procedure to change the Y axis to a maximum value

of 50.


We have a very sharp peak at 0.5 seconds (point 1) where the spring

hits the carriage. Because the peak is so sharp, and the contact force at

this instance qualifies as an impact (or collision) force, we do not know

how accurate this data is. We would need accurate contact elasticity

parameters and more data points to get better accuracy and understand

this impact force.

Just before 2.5 seconds (point 2), the latch reaches the over center point

and we see the maximum contact force of about 36 N. This solution is

reliable and its dependence on the contact parameters is significantly

smaller than at point 1.

1

2

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24 Data points.

Right-click on the curve and click Curve Properties. Select the

Marker tab.

Select Symbol, then OK.


Move you cursor over the data points and the callout will show that the

maximum value is 36 N at 2.42 seconds.

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26 Plot closing torque.

Create a new plot.

Define the plot using Forces, Motor Torque and Magnitude.

Select the RotaryMotor that closes the latch as the Simulation

element.

Again, we observe similar peaks after approximately 2.85 seconds.

These peaks should be ignored for the reason specified in the previous

steps.

27 Modify the plot.

Modify the plot to show the first 3 seconds and a maximum magnitude

of 200 N-mm.


We can see a maximum torque of 96 N-mm at about 2.10 seconds.

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Closing Force Having the torque to rotate the latch, we can determine the force

required by dividing the torque by the distance over which the closing

force acts.

29 Determine the distance.

Click Measure on the Tools

menu.

Measure the distance between the

end of the latch and the axis on

which the motor acts.

30 Required force.

The required force is:

96 N-mm / 25.04 mm = 3.83 N.

Precise Contact Using precise contact instead of tessellated geometry should result in a

more accurate solution, but with a penalty of additional solution time.

We will now solve the problem again with precise contact and compare

the results.

1 Create a new study.

Duplicate the existing study. Right-click the tab for the study

Tessellated geometry and click Duplicate.

Name the new study Precise geometry.

If we experience sudden changes in forces or motions more accurate

solution can be obtained using WSTIFF integrator where the integrator

coefficients are adjusted based on the current step size. The discussion

below describes all integrator types available in SolidWorks Motion

and states when they should be used.

Integrators A set of coupled differential and algebraic equations (DAE) define the

equations of motion of a SolidWorks Motion model. A solution to these

equations is obtained by integrating the differential equations in such a

way that the algebraic constraint equations are also satisfied at every

time step. The speed of the solution depends upon the numerical

stiffness of these equations; the stiffer the equations the slower the

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A set of ordinary differential equations are characterized as numerically

stiff when there is a wide spread between high and low frequency

eigenvalues, with the high-frequency eigenvalues being overdamped.

Special efficient integration methods are required to solve numerically

stiff differential equations because usual methods for solving

differential equations perform poorly and are too slow.

The SolidWorks Motion solver offers three stiff integration methods for

computing motion.

GSTIFF The GSTIFF integration method developed by C. W. Gear is a variable

order, variable step size integration method. It is the default method

used by the SolidWorks Motion solver. The GSTIFF method is a fast

and accurate method for computing displacements for a wide range of

motion analysis problems. For more information on this integrator see

Gear (1971a and 1971b).

WSTIFF WSTIFF is another variable order, variable step size stiff integrator.

Both methods are very similar in formulation and behavior. Both of

them use a backwards difference formulation. The only difference is

that the coefficients used internally by GSTIFF are calculated assuming

a constant step size whereas in WSTIFF, these coefficients are a

function of the step size. So if the step size changes suddenly during

integration, GSTIFF introduces a small error in the solution whereas

WSTIFF can handle it without any loss of accuracy. So the problems

run more smoothly in WSTIFF. Sudden step size changes occur

whenever there are discontinuous forces, discontinuous motions or

abrupt events such as 3D contacts in the model. For more information

on WSTIFF integrator see Van Bokhoven (1975).

SI2 The Stabilized Index Two (SI2) method offered in SolidWorks Motion

is a modification of the GSTIFF integration method. This method

provides better error control over velocity and acceleration terms in the

equations of motion.

Provided the motion is sufficiently smooth, SI2 velocity and

acceleration results are more accurate than those computed with

GSTIFF or WSTIFF, even for motions with high frequency

oscillations. SI2 is also more accurate with smaller step sizes, but is

significantly slower. For more information see Brenan et. al. (1996) and

Gear et. al (1985).

All references are listed at the end of this lesson. For more information

please see Appendix A.PRE-

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2 Change study Properties.

Click Motion Study Properties on the MotionManager toolbar.

Select Use Precise Contact.

Click Advanced Option. Select WSTIFF integrator and reduce the

Maximum Integrator Step Size to 0.0005.

Click OK.


This simulation will take longer to run and depending on your

computer may be around 7 minutes.

4 Contact forces.

Create a plot for the contact forces between the latch and the keeper.

5 Examine the plot.

The plot is similar to the plot we obtained with the tessellated geometry

except that the area where significant oscillations and peaks are

present.

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6 Modify the plot.

Modify the plot to show 3 seconds of time and a maximum force of

50 N.

Within the area of interest, we essentially have the same plot as

obtained with the tessellated data. The maximum force is again 36 N at

2.42 seconds.

We can therefore conclude that we did not need precise geometry to get

accurate results.


Summary In this lesson we analyzed a closing and latching operation of the

latching mechanism. The action of a human hand was simulated with a

help of a motor which controlled the motion of the J_Spring

component. The objective of this lesson was to extract the closing force

and obtain the contact force between the Spring Lever and the

Keeper.

The assembly, initially not fully defined, was completed with a help of

additional mates and zero displacement motors. At some occasions it is

beneficial to restrict the motion with a help of a zero displacement

motor rather than an additional mate because no extra degree of

freedom is removed (motor is a force added to the system, mate is a

constraint removing certain degrees of freedom). The magnitude of the

motor closing the latch (i.e. simulating the action of a human hand) was

expressed with a help of an expression containing mathematical

functions. List of all accepted functions was presented; the STEP

function was discussed in detail.PRE-

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This analysis also involved high velocity contact of solid bodies. Both

available impact models, Poisson and Impact force models, were

discussed in detail. The accuracy of the contact characteristics

(parameters and geometrical description) as well as the accuracy of

some of the resulting quantities, namely contact forces and

accelerations of the impacting regions, was discussed in detail as well.

The study was run using both available geometrical description

models: tessellated and precise geometry. Several convergence issues

were presented and their solution was shown. Precise geometry study

also introduces various numerical integrators available in SolidWorks

Motion simulation. Alternative WSTIFF integrator was also used to

solve this part of the problem.

Discussion: References

Gear, C.W. (1971a). The Simultaneous Solution of Differential

Algebraic Systems. IEEE Transactions on Circuit Theory, CT-18, No.

1, 89-95.

Gear, C.W. (1971b). Numerical Initial Value Problems in Ordinary

Differential Equations. New Jersey, Prentice-Hall.

Van Bokhoven, W.M.G. (1975, February). Linear Implicit

Differentiation Formulas of Variable Step and Order. IEEE

Transactions on Circuits and Systems, 22 (2).

Brenan, K.E., Campbell, S.I. and Perzold, L.R. (1996). Numerical

Solution of Initial Value Problems in Differential-Algebraic Equations,

Classics in Applied Mathematics. ISBN: 0-89871-353-6 (pkb.).

Gear, C.W., Leimkuhler, B. and Gupta, G.K. Automatic Integration of

Euler-Lagrange Equations with Constants. Journal of Computation and

Applied Mathematics, 12 & 13, pp. 79-90, North-Holland: 1985.

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SolidWorks 2011 Exercise 6Hatchback

129

Exercise 6:Hatchback

A number of contemporary car models are designed as hatchbacks.

Similar to station wagons but smaller in size, hatchback cars allow for

cargo to be loaded into the back of the car, and typically the rear seat

folds down to increase the cargo area.


� Contact Forces on page 100.

� Contact: Solid Bodies on page 109.

� Motor Input and Force Input Types on page 103.

� Modifying Result Plots on page 119.

Project Description

Key to the hatchback car's functionality is the hatchback door itself.

These doors are attached to the car via an upward swinging hinge and

are both supported and assisted by gas pistons. To achieve the same

result in SolidWorks Motion, we will apply a motor to the assembly.

Determine the force exerted by the gas pistons on the door.


Open hatchback from the Lesson04\Exercises folder.

2 Verify units.


3 Create a new Motion Study.

Name the new study Hatchback Steel and set the Type of Study to

Motion Analysis.

Note We will be using reference points. In order to use reference points,

make sure that all components are resolved.

4 Apply Gravity to the assembly.

Apply gravity in the negative Y direction. PRE-

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Exercise 6 SolidWorks 2011Hatchback

130

5 Apply Force to the assembly.

The pressure in the piston will be simulated using Action only force

acting on the piston. (It is therefore assumed that the piston force

remains constant as the piston opens). We will begin with the force

definition of the Left_Cylinder.

Apply a 420 N Linear, Action only force as shown below. Make sure

that the force is applied at the indicated point and its direction is

referenced with respect to the cylinder. This way the force will be

always directed along the axis of the rotating piston.

Under Force Function, make sure the Constant button is selected and

enter 420 N in the F1 field.

Click OK.

Note Make sure that the force is oriented as shown in the figure.

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6 Repeat.

Repeat step 5 for the Right_Cylinder.

Mass Properties On occasions, mass properties of the SolidWorks parts may be

modified. This should, however, be an exception rather than a frequent

task as most of the SolidWorks parts reflect the design intent and their

mass properties are computed automatically.

When mass properties are discreetly assigned, they override the

properties associated with the material specifically applied to the

component.

7 Adjust Mass Properties of Lid-1.

Under Tools, select Mass Properties. The Mass Properties window

will appear.

In the Selected items field, right-click and select Clear Selections.

In the assembly view window, click Lid-1 as shown below.

Select Assigned mass properties.

In the Mass field, enter 13000 grams.

Click OK.

8 Adjust duration of the simulation.

Set the study duration to 2 second.

9 Set the study properties.

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10 Contacts - left side.

Define contact conditions between the Left_Cylinder-1 and

Left_Piston-1. For both materials select Steel (Dry).

Keep all other contact options at their default values.

11 Contacts - right side.

Repeat for the opposite side of the assembly, creating contacts for

Right_Cylinder-1 and Right_Piston-1.


In the SolidWorks Motion Manager, click Calculate .

The hatchback assembly will open correctly.

13 Graph the cylinder position.

Create a Y Component plot of the Center of Mass Position of the

Left_Cylinder-1.


Notice that because the plot is created by default in the global

coordinate system, the initial Y value is -63 mm and the final Y value is

289 mm. We can also observe that the initial collision occurs at

approximately 0.83 seconds, while the assembly has completely

opened and stopped moving at approximately 1.1 second.

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15 Contact force.

As stated in the lesson discussions, with generic contact parameters the

contact force solution will be approximate.

Create a new plot for the magnitude of the contact force between the

Piston and the Cylinder (you may use either of the two pairs since the

assembly is symmetrical).


The two spikes in the graph indicate the initial and the secondary

collision between the piston and the cylinder.

The force magnitudes (22,503 N and 4,210 N at the two peaks)

represent the contact forces at the instant of the collision and have to be

understood as approximate due to the quality of the contact input

characteristics. We can further observe that as the motion ceases, the

contact force reaches a constant static value. To determine the contact

value the limits of the graph need to be modified.

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17 Modify the plot format.

Modify the plot of the contact force so that the static value can be read

conveniently.

We can observe that, when the motion ceases, the static equilibrium is

reached and the contact force is at that stage approximately 367 N. The

accuracy of the static solution is not affected by the selection of the

impact model, nor by the selection of the impact model parameters. We

can therefore conclude that the static solution is accurate.

It was mentioned already a couple of times that the contact Elastic

properties significantly effect the resulting impact contact forces and

accelerations of the impacting region. In most scenarios only

approximate characteristics are available and as a consequence the

resulting impact forces as well as kinematic characteristics of the

impacting objects are approximate. We will now modify the contact

elastic properties and study their effect on the solution.

18 Copy study.

Copy the study Hatchback Steel into a new study called Hatchback Aluminum.

19 Change contact material.

Change the contact material for both contacts to Aluminum (Dry).

20 Run the study.

21 The cylinder position.

Create an identical plot of the Y Component of the Center of Matt

Position.

While the minimum and maximum positions are identical and the

general shape of the graph is very similar, notice that the assembly

stops moving at somewhat later time of 1.15 seconds (as opposed to

1.1 seconds when the material specification was Steel (Dry)).

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Because the contact elastic properties have effect on the accelerations

of the impacting regions as well as on the amount of the energy

dissipated during the collision, the resulting velocities after the initial

impact will be different. The assembly will, therefore, cease to move at

a different (now later) time.

22 Contact force.

Create an identical plot of the contact force.

The maximums at the two peaks are again different, 13,412 N and

2,727 N, respectively. But, the absolute values can not be relied on.

As expected, however, the static force magnitude after the motion

ceases is nearly identical to the solution obtained in the previous study,

367 N.

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23 New study (optional).

Repeat the above procedure and change the contact properties to

Rubber (Dry).

Examine the results and notice that this is an unrealistic scenario. You

will have to extend the length of the study to 20 seconds to reach a

point where the motion ceases. The Lid will bounce many times before

eventually coming to rest.

The static value of the contact force, 376 N, is again very close to the

previous solutions.


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Summary In this exercise we analyzed the opening of a vehicle hatchback. While

in reality the two pistons may generate non-constant and non-linear

force, we simplified the simulation and applied a constant piston force

only. While the force magnitude and its dependence on the piston

position can, of course, be modified in a complex way, it was not the

objective of this exercise.

The final phase of the hatchback opening is when the piston contacts

the back side of the cylinder. We used solid body contact and studied

the hatchback opening characteristics (such as opening time, contact

forces, etc.) as functions of the contact specifications. It was found that

with various specifications the hatchback stops moving at different

times. The last study went to an extreme when we used unrealistic

contact specifications: rubber on rubber. In this situation the hatchback

exhibited large repeated oscillations which would be undesirable.

The contact force magnitudes were also analyzed. While the peak

magnitudes coinciding with the short duration collisions are not

reliable since they require very precise contact characteristics, the static

contact force after the motion ceases is accurate. This was

demonstrated by a very similar result obtained from all three

simulation.

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Exercise 7 SolidWorks 2011Conveyor Belt (No Friction)

138

Exercise 7:Conveyor Belt (No Friction)

A conveyor, consisting of

segmented panels, is driven

around a track.


� Functional Expressions on page 103.

� Modifying Result Plots on page 119.

Project Description

Our goal is to drive the conveyor at a speed of 0.62 m/sec using a force

that is controlled by a function. In the first part of the exercise we will

move the belt with a controlled force. In the second part the force will

be replaced with a motion on a path.


Open Conveyor_Belt from the Lesson04\Exercises folder.

2 Review the assembly.

The assembly has all the mates needed for the conveyor belt to move

correctly.

There are many CAM mates that create the tangency conditions

between the wheels and the closed loop conveyor paths.

Note SolidWorks Motion also supports other SolidWorks Advanced mates

like the Gear mates and Limit mates.

3 Verify units.

Verify that the document units are set to MKS (meter, kilogram,

second).


Create a new motion study and name it Conveyor.

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SolidWorks 2011 Exercise 7Conveyor Belt (No Friction)

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5 Apply a force.

We will start by creating a

force on Plate-1 simulating a

force applied to push the plates

on the conveyor.

Apply a 100 N Constant,

Action Only, Linear Force on

the Plate-1 indicated in the

following figure. Make sure the

force is oriented as shown and

its direction is referenced with

respect to the same plate (i.e.

the direction of the force must

change as the plate moves

around the guides).


Set the Number of Frames to

100 and select the WSTIFF

integrator.

Note This problem can be conveniently solved using the faster GSTIFF

integrator as well. The WSTIFF integrator is used here only for

practice.


Run the simulation for 5 seconds.

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8 Plot the velocity magnitude of the Plate-1.

The velocity of the conveyor plates is increasing linearly. We now want

to maintain the conveyor plate at a constant speed of 0.62 meter/

second.

What are we going

to do next?

We are going to change the definition of the force so that it varies as a

function of the difference in the current conveyor velocity from our

desired conveyor speed. Based on the speed difference, the magnitude

and the direction of the force changes to accelerate or decelerate the

conveyor based on the following expression:

Force = Gain * (Desired Speed - Current Speed) = Gain * (0.62 -

Current Speed)

When the current speed is less that the desired speed, a positive force is

applied to accelerate the conveyor. If the current speed is greater than

the desired speed, then a negative force is applied to decelerate the

conveyor. The gain value controls the force applied to accelerate or

decelerate the conveyor.

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9 Modify the force.

Edit the force and change it’s magnitude from a constant of 100 N to

the following functional expression:

100*(0.62-{Velocity1})

Note To get the {Velocity1} feature into the Expression field, double-click

the Velocity1 feature in the Motion Study Results list.


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The plot shows that the velocity is being held to 0.62 m/sec but it is

getting there too slowly. We will increase the gain to shorten the time it

takes to reach the target speed.

12 Modify the force.

Edit the force and change the equation to:

500*(0.62-{Velocity1})



This time, the conveyor reaches the target speed by 1 second and it then

holds there as the force varies. The variation of the speed is, however,

significant and not acceptable for the manufacturing operation. We can

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Modify the force again so that the gain is 5000. Then re-run the

analysis.


This time the plot is much smoother.

17 Plot input force.

It can be seen that the force initial magnitude is very high. To accelerate

the conveyor from its initially zero velocity. As the conveyor reaches

the desired velocity of 0.62 m/sec, the force magnitude tends to reduce

to zero.

Alternatively, instead of using the force input the conveyor constant

velocity can be ensured by using a path mate motion. This is shown in

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Path Mate Motor

Path Mate Motor feature prescribed motion of a point along a path. It is

required to create a PathMate in SolidWorks prior to defining the Path

Mate Motor in SolidWorks Motion.

The options of the PathMate in SolidWorks controls the Pitch, Yaw and

Roll rotational degrees of freedom of the point along the path.

18 PathMate.

Choose one of the wheels on

the plate where the driving

force is applied. Delete the

CamMate.

In SolidWorks feature

tree, unsuppress the

Sketch1 feature.

Define a new PathMate

between the center

point of the wheel and

the path defined by

Sketch1.

Keep all PathMate

constraints at their

default values of Free.

Note The Pathmate constraints are set to Free because the mechanism is

fully constraints due the presence of the remaining CamMates features.

Delete CamMate

Sketch1

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19 Path Mate Motor.

Define Path Mate Motor.

For the PathMate field select

the PathMate defined in the

previous step.

Make sure that the orientation

of the motion is the same as is

the orientation of the force

used to drive the belt.

Select Constant Speed and

enter 0.62m/s.

Click OK.

20 Suppress force.

Suppress the force feature. This feature is not needed because the

motion is driven by the motor.


22 Velocity plot.

This time the plot is much smoother.

Notice the oscillatory variation of the velocity. With the constant

velocity of 0.62m/s prescribed in step 19, we would expect the

resulting velocity profile of the plate to remain constant as well. Can

you explain these oscillations?


Motororientation

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Exercise 8 SolidWorks 2011Conveyor Belt (With Friction)

146

Exercise 8:Conveyor Belt (With Friction)

This is the same conveyor

used in the previous

exercise.

We will run the same study,

but this time we will

include friction and

examine the changes in the

forces and velocities.

Project Description


that is controlled by a function.




� Precise Contact on page 123.


Open Conveyor_Belt from the Lesson04\Exercises\Conveyor Belt\with contact folder.

2 Review the assembly.

Examine the mates.

The first coincident mate keeps the top of one of the plate pins in the

same plane as the end plate of the conveyor. This prevents the side to

side motion of the conveyer plates.

There are groups of concentric and coincident mates that hold adjacent

plates together.

The remaining mates are the CAM mates that create the tangency

conditions between the wheels and the closed loop conveyor paths.

Instead of using the CAM mates, we will use solid body contact.

Suppress all the CAM mates.

3 Verify units.

Verify that the document units are MKS.


Create a new motion study and name it Solid body contact.

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SolidWorks 2011 Exercise 8Conveyor Belt (With Friction)

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5 Add Contacts.

Add a solid body contact between each wheel and the side plate on the

left side of the model (the same side where the CAM mates were

applied). There will be 12 contact sets.

Select Steel (Greasy) for the material and keep the default values for

both the static and kinematic friction.

Note We only create contacts on the left side of the assembly. The contacts

could be defined on the opposite side to model the problem more

realistically. However, similarly to the previous study with CAM mates

(where the mates were defined on one side only to avoid redundancies),

we will keep the contacts on one side only. The final resultant contact

forces will have to be then divided by two.

Redundancies will be covered in a later lesson.

6 Add Gravity.

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7 Add a driving force.

Add a constant force of 5,000 N to plate1<3> just as we did in the

previous exercise. We need to add the constant force first so that we can

generate a velocity graph that can be used in the functional expression

to control the force.

We need a relatively large force to get the belt moving. In the previous

exercise, any force would move the conveyor as there was no friction.

8 Local mate.

There are two instances of a part called

plate_adjust_p1 on the bottom of the

conveyor that are used to tension the belt.

Add a Lock mate to keep these two parts

in the same position relative to each other.


This study will be very sensitive to contact accuracy, so we need to use

Precise Contact. Also set the Frames per second to 100 and select

the WSTIFF integrator.

10 Run the study.


11 Play the animation.

Play the animation at 25% speed to see how the belt moves.

12 Plot the results.

Plot the velocity magnitude of the plate1.The velocity does not

increase linearly as before since the friction forces act against the input

force and the motion with the contact is more complex.

Note To speed up the work you may interrupt the computations at any time.

This run is only important to enable us to define a velocity plot used in

the following expression.

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13 Edit the force.

Change the force to a function of velocity using the equation:

5000*(0.62-{Velocity1})

14 Run the study.

15 Examine the velocity plot.

The velocity approaches 0.62, but the variation is too large.

16 Plot the Force magnitude.

Create a new plot using Forces, Reaction Force, Magnitude and then

select Force1 as the Simulation element.

We will get a warning message about redundant constraints, click Yes.

Contrary to the previous study, the force does not go to zero because of

the friction.

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17 Edit the plot.

Make the maximum Y value 1000 so that we can see the oscillations

easier.

18 Increase the force.

Edit the force and increase the gain to 50,000.

50000*(0.62-{Velocity1})

19 Run the study.

20 Examine the plots.

The velocity is now nearly constant at 0.62 m/sec.

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The force variation is similar.


Summary In this exercise we analyzed the motion of the conveyor belt on the

fixed guide plates.

The belt was accelerated by an action only force applied on one of the

plates. The magnitude of the force was controlled with the help of an

expression which included the velocity of the belt as a variable. This

way, the input force was directly dependent on the resulting velocity.

Two approaches were shown: the first study simulated the tangential

contact between the wheels and the guides using the CAM mates. To

reduce the redundancies and to simplify the solution only mates on one

side were included. Therefore, the resulting contact forces would have

to be reduced by half.

To add more realism to the simulation, the second study replaced the

CAM mates with the solid body contact. While this approach allows us

to add friction, the computation took longer. When the desired speed of

0.62 m/sec was achieved the input force never came to zero in order to

overcome the opposing friction forces.

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Lesson 5Curve to Curve Contact


� Understand the definition as well as the description of contacts.

� Use expressions to prescribe the magnitude of forces and motors.

� Analyze some causes of the incorrect solution or a contact solution

failure.

� Use alternative numerical integrators.

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Lesson 5 SolidWorks 2011Curve to Curve Contact

154

Contact Forces The objective of this lesson is to get familiar with the definition of

curve to curve contact. This lesson builds on the knowledge acquired in

the previous lesson where solid to solid contact was treated in detail.

Case Study: Geneva Mechanism

The geneva mechanism

was traditionally used in

the movie projectors

where each frame is

exposed for a certain

fraction of a second. The

mechanism allows for the

transformation of the

continuous rotation of the

drive wheel into the

intermittent rotation of the

driven wheel.

Problem Description

For the geneva mechanism, determine:

� The contact force generated on the driving wheel.

� Time variation of the driven wheel rotation.


Open stargeneva from Lesson05\Case Studies folder.


Both the driving wheel and the driven wheel are connected to the

base with two hinge mates. There is no mate relation between the

wheels - this interaction will be handled with the help of the curve to

curve contact.

3 Verify the units.



Name the study curve to curve contact.



Driven wheel

Driving wheel

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SolidWorks 2011 Lesson 5Curve to Curve Contact

155

Curve to Curve Contact

Curve to curve contact can be defined between two curves, either of

which can form a closed loop or remain open. The curve geometry is

approximated by a discrete set of points. It is possible to specify

whether the contact is persistent, i.e. curves are not allowed to separate,

or intermittent, where separation may occur.

Curve to curve contact supports friction and two contact models,

Restitution coefficient and Impact force, described in detail in the

preceding lesson.

Introducing:

Curve to Curve

Contact

Contact is used to define the way two curves interact. Within the

contact definition, we can control the friction and the elastic properties

between the bodies.

Where to Find It � Click Contact on the MotionManager toolbar. Under Contact

Type click Curves.

5 Driven wheel and driving wheel contact #1.

Specify an intermittent curve to curve contact between the driven wheel and the left knob of the driving wheel.

In the Contact PropertyManager, select Curves under the Contact

Type.

Under Selections click the Selection

Manager button and set it to Standard

Selection.

Select the indicated

curve on the driving wheel as Curve 1.

Switch the Selection Manger to Select Group setting.

Select the indicated curve and the click the

Tangent button. The tangent closed loop

defining the edge of the driven wheel will be

populated.

Click OK in the Selection Manager to end the

selection process.

The second curve will be constructed and

Closed Group will be shown as Curve 2.

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Under Materials specify Steel (Dry) for both components. Make sure

that the Friction with the default values is used.

Make sure that the orientation of the outward normal for the Closed Group in Curve 2 field is as indicated in the figure above. The

orientation of the curve can be changed with the Outward Normal

Direction button .

Click OK to close the Contact PropertyManager.

Note The Curves always touch button must remain unchecked, because the

two curves come into an intermittent contact only.


Following the same procedure

specify an intermittent curve to

curve contact between the indicated

curves.

Use the same contact parameters as

those in the preceding step.

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Continue with the definition of the

intermittent curve to curve contact

between the indicated segment of the

driven wheel and the closed loop

curve of the driving wheel.

Use the same contact specifications

as those used in step 5.

Note Make sure that the curves are property oriented.

8 Driven wheel and driving wheel contact #4 to #6.

Define the intermittent curve to curve contacts between the remaining

three segments of the driven wheel and the closed loop curve of the

driving wheel.

Note The last four contact sets can be defined in various ways, for example

in a single definition between two closed loop curves. While this is also

a valid contact definition, it is preferable to define contacts with simple

curves rather than one very complex curve.

9 Driving motor.

Apply a 360 deg/sec driving Rotary Motor to the driving wheel.PRE-

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10 Motion analysis Properties.

Set the Frames per second to 100.

Important! The 3D Contact Resolution and Use Precise Contact options are

only applicable to the contact between solid bodies.


12 Plot contact forces.

Plot the contact forces between the driven wheel and the left knob of the driving wheel.

Define the plot using Forces, Contact Force

and Magnitude.

For the selection field, select the Curve Contact1 item from the Motion

FeatureManager.

Click OK.

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Similarly to the contact force results in solid bodies, the contact force

for the curve to curve contact exhibits sharp peaks due to the contact

stiffness approximations and they have to be ignored. Nonlinear

dynamics solutions would be required for the accurate collision forces.

Also, changing the limits for the graph will not yield meaningful static

results for the contact force (as was the case in Lesson 4, where static

contact force existed). Try to answer why.

13 Rotation of the driven wheel.Plot the variation of the rotation of the driven wheel in time.

The above plot indicates that the output rotation rate for the driven

wheel is 90 deg/sec, or 360 deg in 4 seconds.

Solid bodies vs. curve to curve contact

Lesson 4 and the current lesson introduced the two contact types

available in SolidWorks Motion: solid bodies contact and curve to

curve contact. The question may arise as to which contact definition to

use when.

Mots of the contact situations are best resolved with the solid bodies

contact type, especially when the solution of the system depends on

external forces acting on the objects (dynamic systems). If the contact

path can be described using closed loop or open curves, curve to curve

contact type may be used. However, if the curves used in the contact

definitions encircle the entire objects, and especially if they are very

complex, solid bodies contact may still be favored. Therefore, the

above problem of the stargeneva mechanism could be solved with the

solid bodies contact definition instead.

In the last part of this lesson we will solve this assembly again with the

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Solid Bodies Contact Solution

In the second part of this lesson the same assembly will be solved with

the solid bodies contact.

1 Solve problem with solid bodies contact.

Solve the simulation again with the solid bodies contact. Specify the

appropriate geometry description for this contact solution.

When finished, compare the curve to curve and solid bodies contact

solutions.


Summary In this lesson we analyzed a stargeneva mechanism. This mechanism

was traditionally used in movie projectors where each frame is exposed

for a certain fraction of a second. The mechanism allows for the

transformation of continuous rotation of the drive wheel into

intermittent rotation of the driven wheel. To achieve this

transformation, a contact between the two wheels must be specified.

The analysis began with the introduction of the curve to curve contact

that was subsequently defined between the various parts of the

assembly. Curve to curve contact allows for the selection of both open

and closed loop curves and features the same contact models as the

solid bodies contact: Restitution coefficient and Impact force, both

described in detail in Lesson 4.

The solution of the contact force and the time variation of the driven

wheel rotation were plotted and discussed. It was demonstrated and

discussed that the contact force solution in the curve to curve contacts

features sharp peaks corresponding to the collision instances. While the

high magnitudes of the collision instances should be ignored, the static

(or non-collision) contact forces can always be extracted.

Finally, the differences and the proper usage of the solid bodies contact

and curve to curve contact was discussed. The problem was solved

once more with the solid bodies contact and the solution were

compared.

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SolidWorks 2011 Exercise 9Conveyor Belt (Curve to curve contact with friction)

161

Exercise 9:Conveyor Belt (Curve to curve contact with friction)

This is the same conveyor

used in Exercise 7:

Conveyor Belt (No

Friction) on page 138 and

Exercise 8: Conveyor Belt

(With Friction) on page 146

where solid body contact

was used.




� Precise Contact on page 123.

Project Description

In this exercise the solid body contact will be replaces with the curve to

curve contact and the results will be compared.


that is controlled by a function.


Open Conveyor_Belt from the Lesson05\Exercises folder.

This assembly contains completed file set from the Exercise 8:

Conveyor Belt (With Friction) on page 146, where the solid body

contact was used to simulate the CAM tangent conditions.

2 Motion study.

Duplicate the Solid body contact study into a new study named

curve to curve contact.

3 Delete all solid body contacts.

4 Curve to curve contacts.

Add a curve to curve contact between the edge curve of each wheel and the edge curve of the conveyor_path on the left side of the model

(the same side where the solid body contacts were deleted in the

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Exercise 9 SolidWorks 2011Conveyor Belt (Curve to curve contact with friction)

162

Select Steel (Greasy) for the material and keep the default values for

both the static and kinematic friction.

Check Curves always touch check box.

Note The Outward Normal Direction is not shown because Curves always

touch check box was activated.


Set the Frames per second to 100 and select the GSTIFF integrator

from the Advanced Options.

Check the Replace redundant mates with

bushings checkbox.

Note The Replace redundant mates with bushings option is used in this model due to the complex redundancies situation. Both this option as well as redundancies are subject of Lesson 8.

6 Run the study.


7 Play the animation.

Play the animation at 25% speed to see how the belt moves.PRE-

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SolidWorks 2011 Exercise 9Conveyor Belt (Curve to curve contact with friction)

163

8 Plot the results.

Plot the velocity magnitude of the plate1.The velocity does not

increase linearly as before since the friction forces act against the input

force and the motion with the contact is more complex.

Comparing the above variation of the velocity with the results of

Exercise 8: Conveyor Belt (With Friction) on page 146, we can

conclude that they are very similar. Both show nearly constant velocity

of 0.62 m/sec.


Summary In this exercise we analyzed the motion of the conveyor belt on the

fixed guide plates. While in Exercise 8: Conveyor Belt (With Friction)

on page 146 this problem was solved with the help of the solid bodies

contact, in the this exercise curve to curve contact was used instead. It

was shown that both approaches provide similar results.

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Exercise 9 SolidWorks 2011Conveyor Belt (Curve to curve contact with friction)

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Lesson 6CAM Synthesis


� Use of a spline curve to control the motor.

� Create a trace path of a point to get the CAM profile.

� Create a SolidWorks part with this CAM profile.

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Lesson 6 SolidWorks 2011CAM Synthesis

166

CAMs SolidWorks Motion can be used to create CAM profiles based on

tabular data or input function such as STEP function. We can work

backward by driving the follower with the desired motion, then use the

motion of the follower to create the CAM profile.

Case Study: CAM Synthesis

In this case study we will generate a CAM

profile based on an input follower

displacement from a data set.

Problem Description

Create a CAM that will move the follower based on the following

curve.

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SolidWorks 2011 Lesson 6CAM Synthesis

167


To create the CAM, we will follow the steps below:

� Define the motion of the follower.

This can be done from a table of values and drive the follower

through a motor.

� Create a Trace Path.

The trace path will be in the exact shape of the CAM surface.

� Export the curve to SolidWorks as a sketch.

The trace path can be imported into SolidWorks as a curve and used

in a sketch.

� Extrude the sketch to create the CAM.

1 Open the assembly file. Cam Synthesis.sldasm.

Open Cam Synthesis located in the Lesson06\Case Studies folder.

The assembly consists of a undefined CAM and a

follower.


Verify that the units are set to MMGS (millimeter, gram,

second).


Generating a CAM Profile

To generate a CAM profile, the follower motion is prescribed to the

path profile while the CAM component rotates 360°. Both are specified

in the next two steps.

4 Define a motor to drive the CAM.

Add a rotary motor to drive the CAM shaft at a constant

speed of 120 deg/sec. This will rotate the CAM once

every 3 seconds.

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5 Examine the profile data.

Open the file CAM Input.xls located in the Lesson06\Case Studies\CAM Synthesis folder.

Part of the file is shown at right. It consists of X and Y

coordinates for the position of the CAM follower.

The file also contains a plot of the CAM profile based

on the tabular data. Review it, then close the file.

6 Define a motor to drive the Follower.

Add a linear motor to the top face of the

Follower_Guide. Make sure the direction is as

shown in the image.

Select Data Points to open the Function Builder

window.

Select Displacement for Value (y), Time for

Independent variable (x) and Akima for the

Interpolation type.

Click Import Data and select the CAM Input.csv

file. This file contains just the X and Y data that was

in the Excel file.

7 Add gravity.


8 Simulation Study Properties.

Change the study properties to save 100 Frames per second.

9 Run the study.


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Trace Path SolidWorks Motion allows you to graphically display the path that any

point on a moving part follows. This is called a Trace Path and it was

already used once in Exercise 3: Trace Path on page 71. In this lesson

we will use it to generate a profile of a CAM.

You can select the part that will be used to generate the trace curve by

selecting it in the box labelled Select Trace Point Component.

Optionally, you can select a reference component that defines a

reference frame for the trace path. The default reference frame is the

global reference frame defined by the global coordinate system.

Where to Find It � Create a new plot and select Displacement/Velocity/

Acceleration, then Trace Path.

This field enables you to select a face, edge or a

vertex to define a point generating the trace.

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10 Create a trace path which will define the CAM profile.

Click Results and Plots on the MotionManager toolbar.

Select Displacement/Velocity/Acceleration, then Trace Path.

Select the vertex on the Follower-1 to define the CAM profile and

the surface of the cam to define

the reference component.

Leave the Component to define

XYZ directions empty.

Click OK. to show the trace.

Notice how a CAM profile is generated. We will now

copy this trace path curve directly onto the

SolidWorks part from SolidWorks Motion.

Exporting Trace Path Curves

Now that we have the shape of the CAM, we can use this path in

SolidWorks to create the CAM itself. The trace path curve can be

exported to a SolidWorks part.

Introducing: Create

Curve From Trace

Path

The Trace Path curve can be used to create a curve in a SolidWorks part

to create geometry. This can be done in two ways:

� Create curve from path in reference part.

A part already exists, so the trace path curve can be imported to the

existing part.

� Create curve from path in new part.

If a part has not been created, it can be done directly using this

command.

Where to Find It � In the MotionStudy tree, right-click a Trace Path plot under the

Results folder and select Create curve from trace path.

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11 Copy trace path curve to SolidWorks Part.

Right-click the Trace Path plot under the Results folder and click

Create curve from trace path, then Create curve from path in

reference part.

12 Open the CAM part.

Open the CAM part in its own window.

The curve has been inserted into the part as

a new feature.

13 Extrude the profile.

Create a new sketch on the Front plane.

In the SolidWorks FeatureManager,

select Curve1.

Click Convert Entities on the Sketch

toolbar to project the curve onto the

sketch plane.

Also select the outer cylindrical edge of

the CAM profile and use Convert

Entities to project this edge into the active sketch.

Extrude the sketch to a mid-plane depth of 50 mm.

Make sure that the Merge results checkbox is unchecked.

14 Save and close the part.

Return to the main assembly.

In the last part of this lesson, we will re-run the simulation with the 3D

Contact and verify that the cam profile was generated correctly.

We will need to create solid body contact between the follower and the

cam, and drive the motion with the rotary motor on the cam and turn off

the linear motor on the follower.

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15 Add solid body contact.

Add Solid Bodies Contact between the follower and the cam.

Specify Steel (Greasy) for both materials. Clear Friction.

16 Remove the motion for the follower.

Right-click LinearMotor1 and click Suppress.

17 Add gravity.


18 Motion Study properties.

In the Motion Study Properties, select Use Precise Contact.

Whenever we have point contact, we should use precise contact.


Notice how the follower traverses vertically based on the CAM profile.

20 Examine the motion.

Change to the Back view.

The image is at 1.7 seconds. Notice that

the follower is not touching the cam. This

separation is the result of the momentum

of the follower. Just prior to this time, the

follower was driven up by cam. The cam

profile requires the follower to change

direction rapidly, however the only thing

holding the follower in contact is gravity.

Is this a problem? Probably not as the follower will eventually have

additional components on top of it to force contact with the cam.

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21 Plot the vertical displacement of the follower-1.

Create a plot of the Y Component displacement of the center of mass

of the follower and compare it to the plot in the Excel file. For clarity,

the Excel plot has been inverted. Both plots have the same shape.

Cycle based motion

In machine design the independent variable TIME is often not the most

convenient choice. It may be more comfortable to design all tasks in

terms of one master cycle. Typically, the duration of the master cycle is

set to 360 degrees.

Introducing: Cycle

Based Motion

Cycle based motion allows user to easily modify the duration of the

action, or productivity, in the machine design.

Where to Find It � In the FunctionBuilder window set the input type to Variables and PRE-

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Constants and select CycleAngle.

The duration of the cycle is then specified in

the Motion Study Properties.

22 Edit rotary motor.

Under Motor Type select Segments to open the Function Builder.

In the Function Builder dialog, make sure that the Segments button is

selected.

Keep Displacements for Value (y) and set the Independent variable

(x) to Cycle Angle.

Add a row and enter 0deg and 360deg cycle angle for the Start X and

End X columns, respectively.

Enter 360deg for the final value of the rotational displacement.

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The four graphs indicate the linear increase of the displacement,

constant velocity and zero acceleration and jerk.

The 360 degree rotation in 360 degree cycle angle indicates one

revolution per output cycle.

Note The duration of the cycle angle (or output cycle) will be specified in the

next step.

Click OK to close the Function Builder.

Click OK to save the new definition of the Motor.

23 Study properties.

Set the Cycle time to 3s.


Notice that the resulting motion of the follower-1 is the same as in the

step 21. This is to be expected as both simulation are identical, the

former solved using time as independent variable, the later one then

using cycle angle as independent variable.

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25 Analyze results.

Notice that the resulting motion of follower-1 is the same as in the step

21. This is to be expected as both simulations are identical the

definition of the independent variable. The former one solved the

simulation using time as the independent variable, the later one then

used cycle angle.

26 Adjust the cycle time to 1.5s.


28 Analyze results.

Notice the cam now rotates twice in 3 seconds (study duration).

However, reviewing the trace path we see

that follower-1 detaches from the cam -

this is unacceptable. The cycle time of 1.5

seconds is therefore too small for this

mechanism.

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29 Save and close the file.PR

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SolidWorks 2011 Exercise 10Desmodromic CAM

179

Exercise 10:Desmodromic CAM

The mechanisms can be

actuated and controlled in

various directions using

various mechanisms. One

conventional solution is using

springs to return the

mechanism to the original

position (i.e. valve springs in

engines). An alternative

solution may be a system of

cams called desmodromic cams.

In the following exercise, we will build a simple mechanism using a

traditional torsional spring first. Then we will build a second cam

replacing the torsional spring in the system. This way the mechanism

will be driven using a system of cams only.


� see Generating a CAM Profile on page 167.

� see Trace Path on page 169.

� see Create Curve From Trace Path on page 170.

Project Description

In this project, we have already designed a cam that will drive the link

in a predictable motion. As the cam rotates, it will push the link

counterclockwise through contact. As the cam continues to rotate,

some force is required to have the link follower stay in contact with the

cam. In the first part of the exercise, we will apply a torsional spring to

the link to keep it in contact.

Spring

Separation ifno return force

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Exercise 10 SolidWorks 2011Desmodromic CAM

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Open Desmodromic CAM from the Lesson06\Exercises folder.

The first cam (cam1) is already created and mated to the follower

roller1 with a cam mate.

2 Units.

Confirm that the assembly is set to use MMGS units.

3 New study.

Create a new motion study.

4 Restrict axial motion.

The shaft is currently free to move in an axial

direction. Add a linear motor to prevent any axial

movement of the shaft.

Set the Duration time to 10 s.

5 Add rotary motion.

Add a rotary motor to the shaft to have it rotate

360 degrees in 10 seconds.

6 Cam mate.

Examine the mates in SolidWorks and notice that

there is a cam mate between cam1 and the cam

follower (roller<1>). While this mate is

acceptable for animation, it is unrealistic for

analysis because it forces the two surfaces to stay

together even if they would not in reality.

7 Run study.

Set the study length to 10 seconds and run it. The study will run and

show the motion we desire.

8 Remove the cam mate.

In the FeatureManager design tree, suppress the cam mate.

Note You must return the timeline to zero before suppressing the mate.

9 Run study.

The cam1 still rotates, but the link does not move because there is no

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10 Add a spring.

Explode the assembly to

make it easier to select

the correct surface on the

link.

Add a torsional spring to

hold cams together.

Use a Spring Constant

of 10 N-mm/deg, and 30

degrees for Free Angle.

The direction should be

clockwise when viewed

in the Front view.

11 Add contact.

Apply solid body contact between cam1 and the upper follower

roller<1>. For Specify Material select Steel (Greasy) and select

Friction.

12 Run the study.

The motion is correct and the design works well at slow speeds.

If we run this system at higher speeds, we could run into a problem

where the spring cannot keep the follower in contact with the cam. If

we get separation, we could then run into additional problems with the

follower bouncing of the cam and getting a motion other than that

which we were trying to design.

To force contact, we will design a second cam. When our system is

viewed from the Front view, our first cam was able to rotate the link

counterclockwise through contact, but clockwise motion depended on

the spring. In the next part of this lesson, we will replace the spring

with a second cam which will be able to rotate the link in the clockwise

direction. The two cams work together to maintain positive contact

between the cams and followers.

13 Suppress the torsional spring.

Note You must return the timeline to zero before suppressing the spring.

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14 Delete contact and unsuppress the cam mate.

We are going use the Trace Path function to generate our second cam

path. As we need to generate a path that maintains contact throughout

the full rotation, we will use the cam mate to force the contact.

Delete the contact between cam1 and its follower roller<1>.

In the FeatureManager design tree, unsuppress the cam mate.

15 Run the study.

16 Trace Plot.

Create a new plot to generate the curve of the second cam.

We need to select the center point of the second follower roller. We

can do this by selecting the edge of the second follower roller which

will define the center point. Also select the face of cam2.


We now have the basic

path, but it is too large

because we had to trace

the center of the second

follower roller<2>.

Measure the second

follower roller<2>. As it

is 52 mm, we will have to

reduce the size of the

cam2 by half of this, or

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18 Export the curve in the reference part.

19 Open the part.

Open the part cam2 in its own window.

20 Extrude the new cam.

Create a sketch on the Front plane of

the part.

Use Convert Entity to create a circle

in the sketch based on the outer edge

of the existing part.

Use Convert entity again to create a

curve from the trace. Set the type of

this converted curve to For

construction.

Select the For construction curve

from the trace plot and create an Offset Curve, 26 mm to the inside.

Extrude the new cam2 a depth of 10 mm so that the two solids

coincide. Merge the results.

21 Motion Study.

Return to the assembly window.

We will now run the study using the two cams to

drive the motion.

Suppress the cam mate.

Add contact between each of the cams and its

respective follower.

User Steel (Greasy) for the material and select

Friction.

22 Run the study.

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Both cams stay in contact with their rollers throughout the rotation as

one takes care of counterclockwise rotation of the link and the other

controls clockwise rotation.

Tip Use a vertical split screen to be able to watch both the Front and Back

views as the shaft rotates.


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SolidWorks 2011 Exercise 11Rocker CAM Profile

185

Exercise 11:Rocker CAM Profile

In this exercise we will create a

multi-piece cam that is used to

control the motion of a slider.

The toothed wheel rotates and has attached to it a drive plate and guides

for the slider.

The roller will ride in a path between two

stationary cam plates. This system uses the

inner cam to move the slider radially

outward and the outer cam to move the

slider radially inward.

This exercise reinforces the following

skills:

� see Generating a CAM Profile on

page 167.

� see Trace Path on page 169.

� see Create Curve From Trace Path on

page 170.

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Exercise 11 SolidWorks 2011Rocker CAM Profile

186

Project Description

The assembly rotates at 8,000 deg/sec. On each rotation, the rocker will

move radially based on a predefined schedule which is provided in an

attached file.

Create the cams from the existing parts based on a predefined motion

path provided in the separate file.


Open rocker cam profile exercise from the Lesson06\Exercises

folder.


If we hide the toothed wheel and

drive_plate assembly. We can see that

the two cam plates are in place, but the

cam paths have not been defined.

3 Units.

Confirm that the assembly is set to use

MMGS units.

4 New study.


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5 Define the rocker motion.

Add a linear motor to the bottom face of the rocker.

This motion must be specified relative to another component, so select

the guide plate (699-0431) shown.

Use Data Points, Displacement and load the file Slide Translation Motion.csv. For Interpolation type, select Cubic.

Make sure that the direction is radially outward.

Note You may hide the Plate CAM Assembly for easier definition.

6 Define the rotation.

Add a rotary motor to the drive_plate assembly (or part 699-0414).

Set the motor to rotate at a constant speed of 8,000 deg/sec. The

direction should be counterclockwise when viewed from the Top view.


As the time of the simulation is very short, we will need a high frame

rate to have sufficient points to get a smooth result.

Set the Motion Study Properties to capture 2,500 frames per second.

8 Run the study.

Set the length to 0.045 seconds. This will be one full revolution of the

assembly at the speed of 8,000 deg/sec. The assembly should make one

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9 Define a result plot.

Create a trace path of the center of the

roller (699-0413) on the rocker.

Note If the curve does not look smooth, increase the image quality in

SolidWorks Tools, Options.

10 Create curve.

With nothing selected, right-click the Trace Path plot and select Create

curve from trace path, and then Create curve from path.

Because we have nothing selected, this curve will be a feature in the

assembly FeatureManager design tree.

11 Model.

We are now going to work on individual components of the assembly,

so we do not want to be in the Motion Study.

Click the Model tab.

12 Hide components.

We will be creating the cam paths while in the assembly, so it will be

easier to see what we are doing if parts that are not affected are hidden.

Hide the toothed wheel, Slide Assembly and drive plate assembly.

13 Edit part.

Select the part 699-0416 in the Plate CAM Assembly and click Edit

Part .

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14 Edit sketch.

Edit Sketch3 under Base-Extrude.

This is just a circular sketch that defines the outer face of the part. We

will replace this sketch with the trace path curve with an offset for half

the diameter of the roller.

In the FeatureManager design tree,

select the curve (it will be above

the parts and assemblies).

Use Convert Entities to create a

curve from the trace path curve

created in the previous step and set

its property to For construction.

Click Offset Entities and type

6 mm for the offset (half of the

roller diameter). Make sure the

direction of the offset is inside.

Click OK to confirm the offset.

Delete the original circle from the sketch.

Exit the sketch and the part edit

mode. The profile should look

like the image.

15 Outer cam.

Edit the part 699-0417 in the

Plate CAM Assembly.

On the face facing the 699-0416

component (closer face when viewed

using the Bottom view), create a

sketch using the same procedure.

Offset the same curve, but this time

6 mm to the outside.

Extrude a cut to a depth of 8.8 mm.

Exit the part edit mode.

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16 Determine the inner radius.

Measure the distance from the

center of the outer cam plate

to the vertex shown. This is

the same radius that is needed

to create the curve of the

profile on the keeper.

Highlight the distance and

press Ctrl-C to copy this value

to the clipboard as we will

need it in the next step.

17 Show part.

Return to the Edit Assembly mode and Show the part keeper.

This is the keeper that is used to allow access when assembling the

rocker.

18 Edit sketch.

Edit the sketch for Boss-Extrude1.

Double-click the dimension for the radius of

the arc and paste the measured distance from

the clipboard.

19 Examine the completed cams.

Return to the assembly and examine

the cams that have been created.

We should now have a smooth cam

path.

20 New motion study.

Duplicate the existing motion study

into a new one. Name the new study

with contacts.

21 Suppress the linear motor.

Duplicate the existing motion study into a new one. Name the new

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22 Contacts.

Create solid body contacts between all necessary components.

Tip You can conveniently use the contact groups to minimize the number of

definitions.


Activate Use Precise Contact.

24 Calculate the motion study.

25 Analyze the results.

Verify that the designed cam assembly provides the desired motion of

the rocker.


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Lesson 7Flexible Joints


� Learn about Flexible connectors (Bushings).

� Create Advanced Plots.

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Lesson 7 SolidWorks 2011Flexible Joints

194

Flexible Joints In the physical world, nothing is absolutely rigid as materials have the

ability to deform elastically and plastically. To this point in the course,

mates were all simulated as rigid, which is not realistic. In this lesson,

we will start with rigid mates and then make them flexible to more

realistically model them as they would react in the physical world.

Case Study:System with Rigid Joints

A vehicle is being driven on a test track, which has rumble strips that

are 100 mm in height and spaced 2,100 mm apart. The vehicle is

moving at a speed of 60 km/h. A suspension-steering system is set-up

and will be tested for these conditions.

The model is a geometric representation of a short-long arm (SLA)

suspension subsystem with the steering mechanism.

Steering

IntermittentShaft

Base Caps

Strut Upper

Upper Arm

wheel

Strut Lower

Lower Arm

Tie Rod

Body Ground

Steering Rack

Steering Shaft

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SolidWorks 2011 Lesson 7Flexible Joints

195

Problem Description

The goal of this study is to inspect the toe angle that the wheel exhibits

throughout its vertical travel of 100 mm in jounce and rebound. The toe

angle that the wheel exhibits is for the steering wheel angles of 45

degrees, 0 degrees, and -45degrees.

We will first run the study at the three angles with rigid joints. Then we

will change the joints to flexible and run the study again for

comparison.


To analyze the suspension system, we will follow the steps below:

� Create mates.

We will make sure that all the mechanical mates that are required

have been included in the assembly.

� Define the motion.

Add a linear motor that is driven at the frequency that is created by

the vehicle speed and rumble strip spacing.


Create plots of the tire yaw angle versus the vertical displacement.

� Modify the joints.

Change the joints from rigid to flexible.

� Re-run the study.

The results of the study will be compared against the previous study

with rigid joints.


Open Suspension_Steering_System from the Lesson07\Case Studies folder.

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Before we create a motion study, we need to examine the assembly and

determine how the linkages are connected.

In the Mates folder, there is an Angle mate. Examine the mate. This

mate controls the angle of the steering wheel and will be one of our

study parameters as we can use this mate to turn the steering wheel to

specific angles.

Move the tire vertically

and rotate it. Notice that

the lower arm is not

connected to the lower

strut. Also notice that the

tire can turn, even though

the steering wheel doesn’t

because of the mate.

The five Base_Caps (top

of the strut and two per each arm) are fixed and cannot move.

3 Prepare to apply mates.

We are going to add two mates to the assembly, a rack and pinion mate

to connect the steering rack to the steering shaft, and a lock mate to

connect the bottom of the strut to the lower arm.

Important! Before applying these mates, the tire needs to be returned to its zero

position.

Either close the assembly without saving, and then reopen it to return to

the starting point. Alternatively, use Reload to copy the assembly on

disk back into RAM.

4 Attach Base_Caps<5> to Lower_Arm.

In SolidWorks add a Lock mate between the parts Lower_Arm and

Base_Caps<5>.

Now the two parts are rigidly connected to each other.

5 Float Base_Caps<5>.

Base_Caps<5> is Fixed when we open the assembly. Now that we

have created the Lock mate to the Lower_arm, we must remove the

Fixed mate to allow the suspension to move.

Right-click Base_Caps<5> and click Float.

Fixed

Fixed

Not Fixed

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6 Create a Rack and Pinion mate between Steering_Shaft and

Steering_rack.

Add a Rack Pinion mate between Steering_shaft and

Steering_rack which would be connected through a worm gear.

When the Steering part (attached to Steering_Shaft) rotates by

7 degrees, the Steering_rack part travels 1.0 mm.

Select Rack travel/revolution and type 51.43 mm [(360°/7°) x 1 mm/

rotation = 51.43mm/rotation].

Select Reverse to make the direction correct.

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7 Examine Angle mate.

There is an Angle mate on the Steering part. With all the mates

properly applied, the angle mate will be the equivalent of the driver’s

input. To get -45 degrees, input 45 degrees and click Flip dimension.

Set the angel value back to 0 degrees before leaving this step.

Calculation of Wheel Input Motion

A simple harmonic function motion will be imposed on the wheel to simulate this condition. To achieve this, some preliminary calculations are done based on the inputs. For a harmonic function, we need to find the frequency (deg/sec) and the amplitude (which in this example is the 100 mm height of the rumble strip).

Frequency can be computed from the spacing of the rumble strips

(2,100 mm) and the velocity (60 km/h.).

Frequency (Hz) = velocity / spacing = 60 (km/h) / 2,100 (mm) =

16,666.67 (mm/s)/2,100 mm=7.94 Hz.

The peak-to-peak amplitude desired is 100 mm.

8 Create a motion study.


Angle = 0°Angle = +45° Angle = -45°

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9 Create input motion.

We will add a motor to drive the vertical motion of the wheel based on the frequency of the vehicle moving over the rumble strip at the desired speed.

Add a Linear Motor.

Select the vertex in the center of the wheel hub for the position of the motor. For direction, select the Top plane in the part wheel.

Important! You must use the Top Plane in the wheel part and not a plane outside of the part.

Select Oscillating for the motion type. The amplitude is 50 mm (half the height of the rumble strip) and the frequency is 7.94 Hz. Keep 0deg for the Phase Shift.

Click OK.

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10 Create a spring and damper between Strut_Lower and

Strut_Upper.

Define a spring that is attached at the Point at the top of the strut and

the edge at the bottom.

Enter 60.0 N/mm for the Spring Constant and 405 mm for the Free

Length.

Add a linear Damper with Damping Constant of 0.46 N/(mm/s).

For Display, Coil Diameter = 60 mm; Number of Coils = 10; Wire

Diameter = 10 mm.

Click OK.


Set the study properties to record 500 Frames per second.

12 Run the study.

Run the study for 0.12 second which is one cycle at the input frequency

of 7.94 Hz.

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Understanding Toe Angles

When a pair of wheels is set so that their leading edges are pointed

slightly towards each other, the wheel pair is said to have toe-in. If the

leading edges point away from each other, the pair is said to have toe-

out. The amount of toe can be expressed in degrees (from the angle to

which the wheels are out of parallel), or more commonly, as the

difference between the track widths (as measured at the leading and

trailing edges of the tires or wheels). Toe settings affect three major

areas of performance: tire wear, straight-line stability and corner entry

handling characteristics.

The pink arrow denotes the direction of travel of the car.

13 Animate.

Play back the study at slow speed to observe the motion. If you select

Loop , it will continue to play.

14 Plot the pitch.

Create a new plot and select Other quantities, Pitch/Yaw/Roll and

Pitch.

Select the tire face of the part wheel for the part to create results. This

will plot the Pitch at the center of mass for the wheel part.

The toe angle can easily be determined from the plot.

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15 Plot Toe angle vs. wheel height (the Y displacement).

The previous plot is not important to use as what we are really

interested in is the Toe angle as a function of the vertical displacement

of the spindle.

Edit the previous plot. Under Plot Results versus, select New Result,

then Displacement/Velocity/Acceleration, Center of Mass Position,

Y Component.

Select the same face of the wheel.


Because we have rigid joints, we have two lines that fall on top of each

other. One line is the wheel moving up, while the other is the wheel

moving down.

We will now repeat the simulation for two more configurations:

steering angles 45 deg and -45 deg (simulating a left and right turn,

respectively).

17 Change the steering angle to 45 deg.

Set the timeline to zero.

Important! If you do not return the timeline to zero, before editing the mate, the

mate will still be at zero degrees at time zero and will change to 45° at

whichever point the timeline was when the edit was made.

Edit the Angle mate and change it from 0 to 45 deg (to simulate a left

turn).

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The curves are still straight, but the values are slightly different.

20 Change the steering angle to -45 deg.

Set the timeline to zero.

Edit the Angle mate and change it from 45 to -45 deg (to simulate a

right turn).

Again we have the same shape curves, but different values.

From the three graphs shown, notice how the toe angle changes with

the change in the steering angle.

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System with Flexible Joints

In the physical world, nothing is absolutely rigid as materials have the

ability to deform elastically and plastically. In the previous study, the

joints were all simulated as rigid, which is not realistic. In the following

part of the lesson, we will change the joints to be flexible, which will

more realistically model the real world.

Introducing:

Bushings

Bushing objects are added to model flexible mates used on physical

suspensions. Bushing elements allow deformation in a certain degree-

of-freedom that is not accounted for if the attachment is modeled as

rigid. In this lesson, notice how the Lower_Arm is connected to the

Base-Caps with two concentric mates. These two mates could be

replaced with bushings in order to simulate a flexible connection

between the Lower_Arm and Base-Caps.

Typical bushings used in automotive vehicle design consist of steel-on-

steel, Urethane, or Nylon. The stiffness and damping characteristics of

these bushings are measured by SAE testing methods (see Reference 1)

and depend on the type of vehicle (see Reference 2).

Orthotropic bushings can greatly affect the kinematics (camber, toe

angles) and dynamics (joint, shock forces) results of your model when

compared to a rigid connection. In our simulation, we will use isotropic

bushings.

21 Review the mates in the model.

Review the mates associated with the Lower_Arm and the

Upper_Arm parts. Notice how they are connected to the Base-Caps.

The Base-Caps are connected to the automobile frame. There is no

slack in the mates. However, in real life there is some slack or play

between the arms and base-caps. To incorporate this slack, a flexible

connector or, in other words, a bushing will be used.

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22 Create bushings between the Base_Caps

and Arms.

We want to edit the global mates locally, so we

must edit the mates in the SolidWorks

FeatureManager design tree while staying in the

motion study tab.

Locate the four Concentric mates between

Base_Caps 1 through 4 and the upper and lower

arms.

Edit each mate in turn.

� Select the Analysis tab and make the following changes to each

mate:

� Select Bushing.

� Select Isotropic for both Translational and Torsional.

� For Translational, change the Stiffness to 3,500 N/mm, Damping

to 2.63 N-s/mm and Force to 0.

� Leave the Torsional values at their default settings.

Each mate will now have a Bushing

symbol next to the mate type.

23 Steering angle.

Set the steering angle back to 0°.


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25 Plot Toe angle vs. wheel height (the Y displacement).

The steering angle should already be plotted on the screen. We can see

that there is now some slack in the bushings.

The plot below shows the same plot when joints are used instead of

bushings.

Comparing the results, we confirm that the toe angle is different

between the two conditions (bushings vs. joints).

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26 Review Simulation.

Zoom into the location of the Lower_Arm where it connects to the

Base_Caps. Notice how Lower_Arm interacts with the Base_Caps.

There will also be some slack between the moving parts and the

Base_Caps.

27 Obtain results for the two additional configurations.

Obtain the graph of the Toe angle vs. the wheel height for the two

additional configurations: steering angles 45 deg and -45 deg.

Time = 0.0 Time = 0.025 Time = 0.05

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Summary In this lesson, we learned the use of springs, dampers, and bushings in

SolidWorks Motion. We explored several post-processing options to

analyze the rotational displacements of the model. We also studied the

effect of making joints flexible by introducing bushings.

References [1] Adams, Herb, “Chassis Engineering”, The Berkley Publishing

Group, 1993.

[2] Kirschenbaum, Al, “The Official Ford Mustang 5.0”, Bentley

Publishers, 1993.

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Lesson 8Redundancies

Objectives Upon successful completion of this lesson and exercises, you will be

able to:

� Understand Redundancies and how they affect the simulation.

� Use Flexible mates to automatically remove redundancies in a

mechanism.

� Assign the stiffness to each mate individually.

� Understand how to build assemblies without redundancies.

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Lesson 8 SolidWorks 2011Redundancies

210

Redundancies In Lesson 2, 3D Crank Slider Mechanism, we studied the

recommended approach to building assemblies for the kinematic

simulation, i.e. simulation where our main objective is to obtain

displacements, velocities, accelerations, jerks or possibly some reaction

forces. We covered the fact that mates are used to connect the assembly

components and thus constrain the relative motion of a pair of rigid

bodies. Mates therefore determine how the assembly moves and we

also reviewed some of the most common mate types. In the last part of

this lesson we will discuss this topic in greater detail; we will review

how many degrees of freedom do mates constrain and why can this be

important for the solution of our motion simulation. Before we move

ahead, let us review some of the basic terminology and concepts.

Each unconstrained body in space has six degrees of freedom: three

translations and three rotations about X, Y and Z axes. Any rigid body,

i.e. SolidWorks part or rigidly attached parts forming sub-assemblies,

therefore feature all six degrees of freedom. When we use mates to

connect rigid parts or subassemblies together, each mate (or connection

type) removes certain number of degrees of freedom from the system.

Below, we will review the basic mate types and state how many

degrees are removed when two rigid bodies are connected.

The table below lists the most common mates representing some of the

common mechanical connections.

Mate TypeTranslational

DOF removed

Rotational

DOF removed

Total DOF

removed

Hinge mate 3 2 5

Concentric

(2 cylinders)

2 2 4

Concentric

(2 spheres)

3 0 3

Lock mate 3 3 6

Universal mate 3 1 4

Screw mate 2 2 (+1) 5

Point to point

coincident

3 0 3 (this mate is

identical to the

concentric

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SolidWorks 2011 Lesson 8Redundancies

211

The table below list some of the special mates, which do not necessary

represent a real mechanical connection, but do impose a geometric

constraint on the two connected bodies.

A very large number of mates can be listed in the above tables. As you

can see, not only the mate type determines the number of the

constrained degrees of freedom, but also the pair of selected entities is

important.

Lesson 2 recommended that for the models

where kinematic quantities (displacements,

velocities, accelerations etc.) are required, all

mates should, up to the reasonable extent,

represent real mechanical connections. In the

figure to the right, the door is connected with

two hinges. Both hinges should be defined as

hinge mates for the kinematic solution.

As you will soon learn, that due to the

redundancies, this approach is not sufficient

when joint forces are required, or when the part

is to be exported to SolidWorks Simulation for

the stress analysis.

Mate TypeTranslational

DOF removed

Rotational

DOF removed

Total DOF

removed

Point on axis 2 0 2

Parallel (2 planes) 0 2 2

Parallel (2 axes) 0 2 2

Parallel

(axis and plane)

0 1 1

Parallel (2 axes) 0 2 2

Perpendicular

(2 axes)

0 1 1

Perpendicular

(2 planes)

0 1 1

Perpendicular

(axis and plane)

0 2 2

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Based on the number of degrees of freedom, mechanical systems are

divided into two categories:

� Kinematic systems

� Dynamics systems

Kinematic System For a kinematic system, mates and motors fully constrain all the

degrees of freedom on the mechanism. So the position, velocity and

acceleration of each part are fully defined at every time step based upon

the mates and motions applied by motors. Mass and inertia information

is not needed to decide the motion. Such mechanism is said to have

zero degrees of freedom.

For example, consider the Scissor lift

model shown to the right. The motion of

the scissor lift will always be the same

regardless of the mass of the links or

platform, or the weight of people

(external load) standing on the platform.

Only the force required to drive the lift

will change, depending upon a change

in the mass of any component or the

external load. More weight means that

more force is needed to get from height

A to height B.

Dynamic System In a dynamic system, the resulting motion of parts depends upon the

mass of components and the applied forces. If the mass or applied

forces change, then the motion behavior is different. Such a mechanism

is said to have more than zero degrees-of-freedom.

In the mass string example to the

right, depending upon the mass of the

balls, the motion will be different. Or,

if you swing the ball on the left with a

different force, the motion of the balls

will be different.

In summary, the primary difference between kinematic and dynamic

system is that a kinematic system motion is not influenced by the mass

or applied loads, whereas a dynamic system motion is easily influenced

by changing mass and applied loads.

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All of the systems that were analyzed in Lesson 1 to Lesson 7 can be

considered kinematic systems; i.e. given the mates and the applied

motors, the motion of the systems was always determined and unique.

However, as you will soon learn, all of these were also redundant

leading to unique kinematic results (displacements, velocities and

accelerations), but possibly non-unique dynamics results (for example,

joint forces were not computed correctly because no unique solution

existed). Redundant systems, i.e. systems with redundant constraints

(alternatively we may call them over-constrained systems) are subject

of this lesson.

What are redundancies?

Redundancies relate to modeling a real life system as a mathematical

model and are an inherent problem in rigid body motion simulation. It

is very important that you be aware of redundancies and how they can

effect the simulation and results of a mechanism.

At a base level, redundant constraints occur when more than one mate

constrains a specific degree-of-freedom on a part.

Constraints in SolidWorks Motion remove degrees-of-freedom (DOF)

from the system by adding algebraic equations to the governing system

of DAE’s (Differential and Algebraic equations).

Six algebraic equations used by SolidWorks Motion to represent DOF

constrained by mates are as follows:

Equations 1-3 constrain translational DOF while equations 4-6

constrain rotational DOF, where “i” and “j” represent the first and

second parts respectively. The above equations can be understood as

follows:

1. means that the global X-coordinate of the “i” part

must always remain identical to the X-coordinate of the “j” part.

2. means that the global Y-coordinate of the “i” part must

always remain identical to the Y-coordinate of the “j” part.

3. means that the global Z-coordinate of the “i” part must

always remain identical to the Z-coordinate of the “j” part.

Xi Xj– 0=

Yi Yj– 0=

Zi Zj– 0=

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4. means that the Z-axis of the “i” part always remains

perpendicular to the X-axis of the “j” part (which means no rotation

about the common Y-axis).

5. means that the Z-axis of the “i” part always remains

perpendicular to the Y-axis of the “j” part (which means no rotation

about the common X-axis).

6. means that the X-axis of the “i” part always remains

perpendicular to the Y-axis of the “j” part (which means no rotation

about the common Z-axis).

The " " notation in equations 4-6 signifies a dot product operation.

Recall that when the dot product of two vectors is zero, the vectors are

perpendicular.

Each Fixed mate in your model uses six equations (eq. 1-6), while a

Concentric mate (of two spheres) uses three equations (eq 1-3), a

Hinge mate uses five equations (eq. 1-5), etc.

Notice how each of these mates uses equations 1 and 2. Any such

duplication of constrained DOF can lead to over constraining your

system, or introduce what are known as redundant constraint equations.

SolidWorks Motion outputs warning messages to try to help you

understand which equations are redundant and therefore which DOF

are unnecessarily removed. When you have a redundant constraint, you

have two or more mates effectively fighting to control one specific

degree-of-freedom. In simple cases, the solver will automatically

remove a redundant constraint equation to stop the redundancy. In

complex situations it may not remove the correct one for the

mechanism, affecting the original design.

Important! This leads to the simulation still running, but giving the wrong motion

or answer.

Effects of Redundancies

There are two main failures due to redundancies:

� Simulation failure part way through a solution

As the solver progresses through a solution, it continually re-evaluates

redundancies and removes them from the mechanism. Occasionally

during the re-evaluation, different redundant constraints are removed

based on the current positions and orientations. This can potentially

lead to an inconsistent model. Because the solver is unable to

understand the design intent of a mechanism, it can arbitrarily remove

constraints which are mathematically valid, but not valid from a

functional point of view.

� Incorrect force calculation

An example to illustrate this is covered in the next section.

Zi Xj⋅ 0=

Zi Yj⋅ 0=

Xi Yj⋅ 0=

⋅

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How are redundancies removed in the solver?

Before a simulation is actually run, the solver goes through the process

of detecting if the mechanism contains redundancies. If it detects

redundancies, it will try to remove them, and only if successful, will it

continue to run the simulation. At each time step, it continues to re-

evaluate redundancies and removes them as needed.

There is a certain hierarchy by which redundancies are removed. The

solver will remove redundancies based on the following order:

� Rotational Constraint

� Translational Constraint

� Motion Inputs (Motors)

According to this hierarchy, the solver first looks for rotational

constraints that can be removed to eliminate redundancies. If it cannot

remove any rotational constraints, it will then try to remove

translational constraints. If it cannot remove any translational

constraints, it will then try to remove an input motion (as a last resort).

If all these attempts fail, the solver will abort with a message

instructing the user to check for redundant or inconsistent constraints in

the mechanism (or to see if it is in a locked position).

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Case Study:Door Hinges

Let's investigate this removal procedure with the help of a door

mechanism. The most intuitive way to create mechanical connections

consists in recreating the physical reality. For example, when you see a

hinge, you want to model it with a hinge mate. If there are two hinges

on the same part, like this door, and if you place two hinge mates, you

create some redundancies.

Problem Description

We have a simple door

consisting of a door and frame.

The door is connected to the

frame with two hinges.

Determine the forces on the two

hinges as a result of the weight

of the door.


Open door from the Lesson08\Case Studies folder.

2 Float the door.

When the assembly is opened, both components are fixed and have

zero degrees of freedom.

Right-click the door and click Float.

Frame

Door

Hinge

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3 Add Hinge mates.

To make it easier to select faces on the hinge, move the door a small

distance.

Add a Hinge mate between the two halves of the upper hinge.

Note It is not important if the mate is added as local or global.

4 Add another Hinge mate.

Add a second hinge mate to the lower hinge.

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5 Check the door weight.

Check the mass properties of the door part. The door weighs 28,020.63

grams, so the vertical force of the door should be a 274.8 N.


7 Add gravity.


8 Change study properties.

Edit the motion study properties and set Frames

per second to 50.

Make sure that Replace redundant mates with

bushings is cleared. We will discuss this option

later in the lesson.

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Degrees of Freedom Calculation

Let us review how many degrees of freedom (DOF) are currently

restricted by our mates. Because the frame is a fixed body it features

zero DOF. The only floating body in the assembly is the door. Our

mechanism may therefore feature up to 6 degrees of freedom.

The two hinge mates defined in the model are then each constraining 5

DOF.

Total Actual and Estimated DOF

The current DOF count for our system is therefore , i.e. our

mechanism is over-constrained based on the simple DOF count. This

simple count is referred to as approximate (or Gruebler) and is rather

easy to obtain. It could indicate that our mechanism cannot move. It is

obvious, however, that the door is allowed to rotate about the hinges

and, in engineering sense, should not be over-constrained; using this

engineering approach, our mechanism features +1 DOF (rotation about

the hinges). This count, referred to as “actual” is more complex to

obtain than the simple count introduced above.

The number of redundant constraints in the system is therefore

; our system features 5 redundant constraints –

constraints which are not needed from the mathematical point of view

for the door to close and open. Indeed, by removing one of the hinges

the kinematics of the system is unchanged.


Run the simulation for 1 second. There is no movement in this

assembly.

We will now review the number of degrees of freedom as well as the

number of redundancies with the help of functions within SolidWorks

Motion.

Introducing:

Degrees of Freedom

Calculation

Rather than manually calculating the Degrees of Freedom, SolidWorks

Motion can quickly calculate them for us.

Where to Find It � Right-click the local mate group and select Degrees of Freedom.

6 2 5×– 4–=

6 2 5 1–×– 5–=

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10 Use the simulation panel to calculate degrees-of-freedom (DOF).

When the study completes, notice that in the Motion Study

FeatureManager, the mate folder reads Mates (5 Redundancies), just

as we calculated a while ago.

Right click the local

Mates folder and select

Degrees of Freedom to

open the dialog shown

below. We can review the

number of moving

(floating) parts, number of

mates (presented as joints),

number of the estimated

and actual DOF and the

Total number of redundant

constraints.

SolidWorks Motion

calculates five redundant

constraints. The

mechanism is over-

constrained.

As mentioned above, the reason for this is that a second hinge mate is

trying to constrain the same degree-of-freedom as the other hinge mate.

Numerically, one hinge mate is sufficient to simulate the hinge

condition. But this may not be enough, especially when reaction loads

at both the hinges are to be calculated.

In order to obtain a unique solution, the program is forced to remove

the 5 redundant constraints. The selection is made internally without

user intervention; the removed redundant DOF can also be found in the

above list.

We will now review the force solution in the joints to reveal the

consequence of the redundancies.

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11 Plot reaction forces for the Hinge mates.

The weight of the door is approximately 274.8 Newtons. Gravity acts

along the negative global Y direction. The two hinge mates should

share this load equally. Let us verify this.

Create two plots to show the Y Component reaction force for the two

hinges. When we define the plot, we will be warned:

The motion study has redundant constraints which can lead to invalid force results. Would you like to replace redundant constraints with bushings to ensure valid force results? Note that this will make the motion study slower to calculate.

As redundant mates are the subject of this lesson, we will first see what

happens with the redundant constraints.

Click No.

The reaction force on one of

the hinge mates is zero

which should not be the case.

On the other hinge mate the

reaction force is 274.8 N.

It can be seen that one hinge

mate carries the full force of

274.8 N, while the other one

carries no load. The

distribution of the forces

between the two hinges is

incorrect.

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Let us see why the simulation gave us such results. In Step 10, using

the simulation panel, we calculated the DOF in the mechanism. Notice

that one of the redundant constraints was mentioned as “Hinge2: Translation along Y”. This tells us that the mechanism is already

constrained in the Y direction by the Hinge1 mate. The same degree-

of-freedom is being constrained by the Hinge2 mate and will be

ignored. Therefore, no results are calculated for the Y-direction reaction

force on the Hinge2 mate.The entire weight of the door will then

have to be reacted upon at the Hinge1 mate at simulation time.

Likewise results for other redundant constraints will be ignored and

hence turn out to be zero.

We will now see how this issue can be avoided by using the Flexible

joints option.

Using Flexible Joints Option to Remove Redundancies

In the discussion on page 214 it was mentioned that the redundancies

may lead to:

1. Simulation failure part way through a solution, and

2. Incorrect force calculation (distribution).

The effect of point 1 can be minimized (though not avoided) by using

mates closely representing the mechanical connections in the real

product. For example, two hinges in the door-frame assembly could be

mated with two hinge mates since they represent the real connection

type the most closely. Alternatively, point 1 can be tackled by reducing

the number of redundant constraints manually by using simpler mates

such as point on axis and similar. In complex assemblies this can be,

however, daunting task and may require iterative approach of mate

design and the DOF calculation. For example, imagine that in our

current example of the door, one hinge mate is deleted and the number

of redundant constraints is then zero; the solution in the Y direction is

then identical.

The effect of point 2 can be tackled by manually modifying the mates

to remove the redundancies in the requested (or all locations) and re-

adjusting the distribution of the reaction forces in the mates manually,

or using the technique introduced in Lesson 7, flexible mates. To

demonstrate the former, imagine that one redundant hinge mate is

deleted from the simulation – all load is then carried by the remaining

hinge mate. Knowing the geometry, we manually readjust the

distribution equally into both mates. This approach may work in simple

design and loads such as the current example of the door or many

symmetrical mechanisms such as fork lift (analyzed in some of the

exercises following this lesson). In the later approach, when flexible

mates are used in place of mathematically rigid mates, stiffness of the

mates in the respective directions decides on the distribution of the

reaction forces.

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While this approach is still an approximation, it can provide a more

realistic distribution of forces than the infinitely stiff case.

When you make a mate flexible, the mechanism will be updated to

have a bushing representation of the basic mate type instead of a rigid

constraint. Mate motion and friction are not affected by using flexible

mates.

Limitations of Flexible Mates

The following limitations may exist when using flexible mates:

� In some models, using bushings will slow down the solve time

because of induced dynamic effects.

� We are not accounting for the stiffness of the part in the solution.

Therefore, the distribution of loads due to part stiffness may differ

from the bushing constraint solution. This bushing approach will

ensure that force results are obtained at all mate locations. This

limitation, however, exists in the case of rigid mates solutions as

well.

� Advanced mates do not support mate flexibility. See the Help for a

list of joints that can be made flexible.

� If the mechanism starts in a dynamic condition, there may be a

spike in initial forces as the model reaches initial equilibrium (that

you would not see with rigid joints). The spike is generated by

initial conditions of the parts not balancing and the bushings

resisting rapid changes in force/acceleration. If the model started

with enforced motions (e.g., constant velocity), try ramping up

motions from zero to the desired value over a time range to

eliminate or minimize this (e.g., use a step function to ramp

velocity from zero to a certain value over a time range).

� An optimum mate stiffness and damping characteristics may need

to be entered. This may require an iterative approach.

The following joints can be made flexible: Fixed, Revolute,

Translational, Cylindrical, Universal, Spherical, Planar, Orientation, In

Line, Parallel Axis, In Plane, Perpendicular.

In SolidWorks Motion Simulation, the flexibility in the mates can be

introduced in two distinct ways.

1. Replace redundant mates with bushings option in the Motion Study

Properties. This way, one set of global stiffness and damping

characteristics is assigned to some algorithmically selected mates

only. The decision on which mates are made flexible and which are

kept rigid is done by an advanced algorithm and is fully automatic.

This approach may work in most situations.

2. Assigning individual stiffness values to the selected (or all) mates

manually. This technique will work in all situations, but can be time

consuming. Local mates may be used with great advantage without

altering the design intent of the assembly designer.

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Where to Find It � In the Motion Study Properties, select Replace redundant mates

with bushings.

When mates become flexible, the .icon will appear next to the mate

icon in the MotionManager tree.

Bushing Properties

When bushings are defined, their Translational and Rotational Stiffness

and Damping may be defined.

Where to Find It � In the Motion Study Properties, click Bushing Parameters.

In the remainder of this lesson, we will use the Replace redundant

mates with bushings option to correctly solve the door example.

Assigning individual stiffness to the selected mates manually as well as

manual removal of the redundant constraints by building redundancies-

free assembly models are practised in the exercises following this

lesson. Students are encouraged to complete all of the following

exercises to fully understand this important subject.

12 Make joints flexible.

Make all joints in the mechanism flexible, in the Motion Study

Properties, select Replace redundant mates with bushings.

Click Bushing Parameters. If we were to change the stiffness and

damping values of the hinges, we would do it here. To see the effect of

the mates stiffness on the solution, complete the following exercises.

Click OK twice.


Notice how the mate icon in the Mates

folder of the MotionManager changes. The

yellow lightening indicates that the

flexibility of this mate was forced by the

software rather than manually specified by

the user (as was the case in Lesson 7).

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14 Review results.

The Y Component of the

Reaction Force for both

mates now shows a value of

137.5 N.

The weight of the door has

now been correctly shared by

both of the Hinge mates.

Note The approach to make

selected (or all) mates flexible

was introduced and practised

in the previous lesson and will

not be shown here.


Important! Students are encouraged to review the following discussion and the

exercises following this lesson. The subject of redundancies is not

trivial and must be understood in order to correctly conduct dynamic

analyses (analyses where correct force distribution is required).

How to Check For Redundancies

As mentioned previously, it is important to only supply sufficient

constraints to obtain the required motion on any mechanism.

Kinematic/Dynamic analysis needs only the necessary degrees-of-

freedom restrained in a system.

A quick indication of whether a system is over-constrained is the

Gruebler count.

� If the number is greater than zero, then the model is under-

constrained (dynamic).

� If the number is equal to zero, then the model is fully defined

(kinematic).

� If the number is less than zero, the model is over-constrained

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An important aspect behind modeling mechanisms is in recognizing the

restrained freedoms of connecting parts and making sure they are not

repeated. This can be difficult in very complex assemblies, but will

ensure you achieve the desired motion and force results. If this is not

taken into consideration, redundant constraints will have been applied

which may result in the simulation not working.

Typical Redundant Mechanisms

Several mechanisms are redundant by their nature. In the real world,

assembly tolerance, slop, and stiffness make the mechanisms work, but

in the mathematical world, they can be invalid. Below are a couple of

examples of these mechanisms.

Dual Actuators Driving a Part

From a kinematic point of view,

you only require one actuator to

move a part. In the real world,

pairs of actuators are used to

provide balanced loads from side

to side. The main problem in the

motion simulation world is that

motions are enforcing

displacement in a specific

degree-of-freedom. A specific degree-of-freedom is constrained by two

actuators. Thus, by having two motions you are causing a redundancy.

This can lead to two situations. One, that only one actuator carries the

load and the other has no result, or two, that there is artificial load

induced into the system (equal and opposite) that produces incorrect

driving force results on the motions. Ways to work around this problem

are to use non-rigid connections to link each actuator into the

mechanism or to use a force based movement instead of a motion based

movement.

Parallel Linkages The scissors lift is a classic example

of where one complete side of the

mechanism is redundant but is done

to provide balanced loads on both

sides of the structure and to make

the design easier. It is simpler to

work with a lighter strut that only

carries in-plane load than to design a

heavier strut that must not only carry

in-plane, but out-of-plane torsional

loads. However, for mechanisms, it

is easier to model only one side and

let the other side “come along for

the ride”. When analyzing these

types of mechanisms, you can lock

the duplicate parts together by attaching them together or using fixed

joints. You then need to delete any duplicate constraints. When you

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obtain the joint loads, just remember to divide by two. Also, remember

that out-of-plane moments should only be due to the non-symmetry of

modeling one side, and the moment should equate to half of the

reaction force times the distance between the two sides that raise the

platform (see Exercise 14: Kinematic Mechanism on page 233).

Summary In this lesson, we defined and familiarized ourselves with the concept

of redundancies. Redundancies occur when identical degrees-of-

freedom in the assembly are constrained by multiple joints. Models

with such redundant constraints are improperly defined and their

solutions are likely to be incorrect (or impossible to obtain). The

implication of redundancies was demonstrated in the first part of this

lesson.

In reality, the rigidness of joints is only an idealized concept. As such,

SolidWorks Motion enables users to disable such rigidity and specify

some finite stiffness and damping along the constrained degrees-of-

freedom in the joint. This approach eliminates the problem of

redundancies, but introduces additional parameters that must be

specified (joint stiffness and damping). Flexible joints were the subject

of the second part of this lesson.

Students are encouraged to complete the following exercises.

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SolidWorks 2011 Exercise 12Dynamic Systems

229

Exercise 12:Dynamic Systems

This exercise will demonstrate a simple

dynamic system with four spheres dropping

in a closed container.


� Dynamic System on page 212.

Project Description

Four aluminum balls are contained in a closed container and will fall

under gravity. None of the components have mates and are free to

interact with each other. Examine the motion of this dynamic system.


Open Vase with Spheres from the Lesson08\Exercises folder.

2 Create a new motion study.

3 Add gravity.


4 Add contact.

Add contact between all components.

Specify Aluminum (Greasy) for both materials.

Add Friction using the default values.


Run the study for 1 second.


All but one ball falls through the container. This can be cause by either

a very coarse time step or too coarse of a contact description.

7 Change study parameters.

Change the motion study properties to record 600 frames per second to

save more data on the disk and specify Precise contact.

8 Run.

Re-run the simulation.

All four spheres are now contained within the Vase. As the spheres

fall, they interact with each other and the vase.

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Exercise 12 SolidWorks 2011Dynamic Systems

230

9 Animate.

Play the study at 10% speed and examine the motion of the spheres.


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SolidWorks 2011 Exercise 13Dynamic Systems 2

231

Exercise 13:Dynamic Systems 2

This is another exercise with a

dynamic system. In this study we

will compare our hand calculation

of degrees of freedom with those

calculated by SolidWorks Motion.

We will also investigate the

effects of changing the impact

from elastic to plastic.


� Poisson Model (Restitution Coefficient) on page 110.

� Dynamic System on page 212.

Project Description

Five spheres are attached to individual frames. One end sphere is pulled

away from the others and released. Examine the motion of the five

spheres with both elastic and plastic impact.


Open Momentum from the Lesson08\Exercises folder.

2 Calculate the DOF.

Calculate the DOF by hand. The DOF should be a positive number to

confirm it is a dynamic system.

3 Create a new study.

Create a new study and name it Rest Coef = 1.0.

4 Add gravity.


5 Add contact sets.

Add four contact sets between the pairs of balls

that will make impact.

Clear Material and Friction.

For Elastic Properties, select Restitution

coefficient and set it to 1.0 which is an elastic

collision.


Record 25 frames per second.

Select Use Precise contact.

7 Run the study.

Run study for 5 seconds.

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Exercise 13 SolidWorks 2011Dynamic Systems 2

232


We see nearly elastic contact. If it was exactly elastic, we would not see

the interior balls move, however with the slight errors in the numerical

methods used, we see some movement as the study progress.

9 Degrees of freedom.

Now that we have run the study, we

can let SolidWorks Motion calculate

the DOF so we can compare the

results with that which we calculated

by hand.

We have five moving parts with six

degrees of freedom for a total of

thirty. The five hinge mates remove

25 degrees of freedom, leaving us

with five degrees of freedom total.


Name the study Rest Coef=0.1.

11 Edit contacts.

Edit the five contacts to and change the Restitution coefficient to 0.1.

This is nearly plastic impact.

12 Run the study.


With plastic impact, once the first sphere makes contact, all the spheres

move together as we would expect.


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SolidWorks 2011 Exercise 14Kinematic Mechanism

233

Exercise 14:Kinematic Mechanism

The following exercise will

demonstrate a kinematic mechanism.

The basic characteristic of the

kinematic mechanisms is a possibility

of a single motion, irrespective of the

applied forces and motors, contrary to

the dynamics mechanisms

(demonstrated in the previous

exercises) where multiple motions

may exist. The Scissor Lift

demonstrated in the following exercise

features no redundancies and one

“Actual” DOF. We will “consume”

this last DOF on the motor which will

drive the mechanism.


� Kinematic System on page 212.

� How are redundancies removed in the solver? on page 215.

� Effects of Redundancies on page 214.

� We will now review the force solution in the joints to reveal the

consequence of the redundancies. on page 220.

� How to Check For Redundancies on page 225.

Project Description

Analyze the mates utilized to

built this assembly. Notice,

that as suggested in the

discussion on page 222, only

half of the symmetrical

mechanism is used for the

mate definitions. The

symmetrically located

components move in phase

with the mated components.

Also note that the assembly

features many geometric

constraints (non-mechanical

mates such as coincidence of

a point and axis, or

coincidence of two planes). Mates onthis side

No mateson thissidePR

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Exercise 14 SolidWorks 2011Kinematic Mechanism

234

Examine the individual mates,

such as Coincident14. Notice

that this, like many of the mates

in this assembly, is a geometrical

constraint (point and face) rather

than a mechanical mate (hinge).

Use of such mates requires

existence of the reference entities

and the building procedure can

be time consuming; the DOF

count must be checked after each

rigid component is added until

the whole assembly is completed.

Due to the time constraints we

will not build this assembly in its

entirety; only its part showing the procedure is demonstrated in the

following exercise.


Open Scissor_Lift from the Lesson08\Exercises\Kinematic Mechanism folder.


Examine the existing mates and move the assembly. With the existing

mates, the only motion allowed is that which moves the platform

vertically.

3 New motion study.

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4 Add motor.

We will add one Linear Motor to the piston to drive the motion of the

assembly.

Add a linear motor to the piston.

Set the Motion to Oscillating, 100 mm at 0.5 Hz. with 0deg for Phase

Shift.

Set the motion to move relative the cylinder.


Set the study properties to record 50 Frames per second.

6 Run the study.


7 Degrees of freedom.

In the local mate group, right-click

MateGroup1 and select Degrees of

Freedom.

We now have zero Total DOF because

of the addition of the motor.

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Exercise 14 SolidWorks 2011Kinematic Mechanism

236

8 Plot forces.

To see the consequences of

modeling the assembly

with geometric mates on

just one side, we will plot

the forces in two of the

mates, Concentric14 and

Coincident9.

Create plots of the

Z Component of the Reaction Force for each mate in the global

coordinate system.


The plot for mate Coincident9

shows a maximum force of

15,166 N. Because of the

redundancies, this is actually the

combined force on both sides of

the assembly. At this joint, the

real force will be half or about

7,583 N.

For Concentric14, the

maximum combined force is

9,536 N which means that each

side carries 4,768 N.

Note The assumption that each side of

the assembly carries half of the

total force at each joint is based

on symmetrical loading of the

assembly.

10 Plot moments.

Create a plot of the X

Component of the Reaction

Moment for the Tangent3

mate.

This moment about the global X

axis should be zero if both sides

of the assembly carry the load

symmetrically. This moment is

only a product of the way this

assembly was build and will be compensated by the reaction force on

the opposite side.


Concentric14

Coincident9

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Summary This exercise demonstrated a complicated model build without any

redundancies. Such procedure may be time consuming; use of the

geometric constraints such as coincidence of points and axes or planes

is necessary and the step by step procedure with frequent DOF

calculation is necessary. This procedure, applied to a part of the large

assembly (sub-assembly) or the entire assembly may only be required

if all other means of attempting solution failed.

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Exercise 15 SolidWorks 2011Zero Redundancy Model-Part 1

238

Exercise 15:Zero Redundancy Model-Part 1

The following exercise will

demonstrate a short section of

the model building procedure

for the mechanism with zero

redundancies featured in the

previous exercise. We will re-

use the model from the

previous exercise, Kinematic

Mechanism, at an early stage

of the model building phase.

The model will feature one redundancy. The goal of this exercise is to

remove the redundant constraint with the help of multiple geometric

constraint (simple mates such as coincident of point and axis and

similar).


� Redundancies on page 210.




Project Description

The same scissor lift used in the previous exercise will be used to

practice the procedure of removing and controlling the number of

degrees of freedom in the model. We will start with just the base and

first layer of scissors, remaining components have been suppressed.

The components of interest in this exercise will be the cylinder and

piston.


Open Scissor_Lift.sldasm from the Lesson08\Exercises\Zero Redundancy Model folder.

The platform and layers 3 through 6 are suppressed.

2 Run the analysis Exercise Study.

The motor is already set up as it was in the previous exercise, so you

can just click Calculate.

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SolidWorks 2011 Exercise 15Zero Redundancy Model-Part 1

239

3 Examine the Degrees of

Freedom.

The local mate group shows

one redundancy. Right-click

MateGroup1 and click

Degrees of Freedom.

With zero Total DOF, the

mechanism will move as we

expect.

We can also see that there is

one redundant constraint and

that the redundant constraint Concentric16, Rotation about X is removed.

4 Determine orientation.

This redundant constraint is about the

local X axis, so how do we determine

which way the local axes are oriented?

Create a plot based on the mate.

Concentric16 is a mate between the

cylinder and the piston. Create a plot of

the Reaction Force, Y Component of

the mate Concentric16. We do not

actually need to complete the plot, but

once the plot is set up, we can see the

orientation Triad.

The X direction (red) is along the common

axis of the two parts. Once we observe the

direction, cancel the plot.

This concentric mate is redundant because

neither the piston nor cylinder can rotate about this axis. The

cylinder has a hinge mate to connect it to the Base, and the piston

has a concentric mate to the cross rod.

5 Remove mate.

The piston and cylinder need to stay concentric,

so we will not delete that mate. Instead, we will

replace the hinge mate, which removes five

degrees of freedom, with two less complicated

mates.

Delete the mate Hinge1.

The end of the cylinder is now free to move.

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6 Add mate.

For the following mates, we will be mating points

and axes, so make both visible.

Note To see the points you need to set the assembly mode to Resolved.

There are already two points created in the hole at the end of the

cylinder. Point1 is on the axis of the hole, half way between the

parallel faces. Point2 is also on the axis of the hole, but coplanar with

the side face.

Add a Coincident mate

between Point1 and the axis

of the hole in the frame.

Add a second Coincident mate

between Point2 and the inside face

of the bracket on the frame.

Note You can add these mates as either local or global.

7 Run.

Run the simulation and observe the results. The study appears to run

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SolidWorks 2011 Exercise 15Zero Redundancy Model-Part 1

241

8 Check the DOF.

Right-click MateGroup1 and click

Degrees of Freedom.

There is still one actual degree of

freedom, but there should be none for

correct results.

9 Determine the problem.

Try to drag the piston

along the cross_rod.

When you do, you can see

the cylinder rotate.

10 Edit mate.

The mate between

Point2 and the face is

not enough to stop the

rotation, so we will

have to raise the level

of the mate to remove

an addition degree of

freedom.

Edit the Coincident

mate and replace Point2 with the edge shown.

11 Run.

Run the study and observe the results.

Remove point2 Add edge

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12 Check the DOF.

Right-click MateGroup1 and click

Degrees of Freedom.

We now have zero DOF as we should

for a kinematic system.


If you are going on to the next

exercise, leave the assembly open,

otherwise save and close the files.

Summary This exercise demonstrated how a mate with redundant constraint is

detected, removed and replaced with a combination of simpler

geometrical constraints such as coincidence of a point and an axis. As

was mentioned in the previous exercise, this technique requires

additional reference entities (points, axes) and the building procedure

can be lengthy. It should be used if all other techniques failed to give

the desired solution. In general, it is easier for the solver to obtain the

solution for models without redundancies then for models with

multiple redundancies.

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SolidWorks 2011 Exercise 16 Zero Redundancy Model-Part 2 (Optional)

243

Exercise 16:Zero Redundancy Model-Part 2 (Optional)

This exercise continues the process

started in the previous exercise. The

remaining parts and subassemblies

with the mates in the Scissor_Lift assembly need to be add to achieve

zero degrees of freedom for correct

results.


skills:





Project Description

Add mates to the assembly to achieve zero degrees of freedom.


Continue working on the model prepared in the previous exercise. If

the assembly is not open, open Scissor_Lift from the Lesson08\Exercises\Zero Redundancy Model folder. Complete the previous

exercise first and then continue with this exercise.

2 Unsuppress layers.

Unsuppress the sub-assemblies

layer_3 and layer_4.

3 Repair mates.

Mate them to the rest of the assembly

so that the mechanism operates as

required with redundant constraints

and actual DOF count both equal to

zero.

Continue building mates on the left

side as indicated in the figure.

4 Continue.

Unsuppress the sub-assemblies layer_5

and layer_6.

Continue adding mates to achieve zero

degrees of freedom.

5 Continue.

Unsuppress platform.

Continue adding mates to achieve zero

degrees of freedom.


Left side Right side

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Exercise 17 SolidWorks 2011Removing Redundancies with Bushings

244

Exercise 17:Removing Redundancies with Bushings

In this exercise, we will use a model that has mates applied

symmetrically in preparation for exporting results to SolidWorks

Simulation. To remove the redundancies, we will add bushings and

explore the effects of different damping values.




� Using Flexible Joints Option to Remove Redundancies on

page 222.

� Bushing Properties on page 224.

Project Description

This is the same Scissor Lift assembly used in the previous exercises

except that the components are mated differently.


Open Scissor_Lift from the Lesson08\Exercises\Redundancies Removal with Bushings\completed-low stiffness folder.


The approach to the assembly

mates is different in this model.

Notice that most mates, namely the

Concentric mates, represent the real

mechanical connections closely.

Coincident mates only ensure that

the assembly does not move

sideways.

Notice also that, unlike

previous exercises,

mates are applied on

both sides of the

symmetry.

This method is

appropriate when we

want to import the

results in SolidWorks Simulation to get stress results in the different

components, or if we want to see the correct force distribution at all

mate locations on the model.

The problem with this mating scheme however, is that we are going to

generate a large number of redundancies that will have to be removed.

Concentric36 Concentric37

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SolidWorks 2011 Exercise 17Removing Redundancies with Bushings

245

We saw in a previous lesson that we could have SolidWorks Motion

treat all redundant mates as bushings. We can also manually configure

mates to act as bushings.

3 Add bushings manually.

To save time, each concentric mate has already

been configured as a bushing.

Edit one of the Concentric mates.

Select the Analysis tab.

Notice that Bushing has been selected and the

values set as follows:

Translational

� Select Isotropic

� Stiffness = 5,000 N/mm

� Damping = 20.0 N-s/mm

� Force = 0 N-mm

Torsional


� Stiffness = 100 N-mm/deg

� Damping = 20.0 N-mm-s/deg

� Toque = 0.0 N-mm

These values for stiffness and damping are very

low for a practical system, however we will start here to see the effect

on the mechanism.

Notice that each mate that has been defined as a bushing now features

the bushing icon shown in the MateGroups.

4 Run.

Notice that the motion is not smooth.

Play the animation back at a slower speed and watch the action of the

individual joints.

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Plots of the reaction force in the

Z direction have already be

created for mates Concentric20

and Concentric21. These mates

are on opposite sides of the

assembly.

Notice that the plots are exactly

the same.

6 Create additional plots.

Create additional plots of the force

in the Z direction for mates

Concentric36 and

Concentric37. These mates are

between the lower cross arms and

the brackets on the frame.

Note When we create these plots, we still get a warning message about

redundancies. This will be explained in upcoming steps.

Click Yes to dismiss the message.



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As with the previous set of mates these plots are also identical. While

they are identical, they are not the sinusoidal shape that matches the

driving motor due to the low stiffness of the joints.

Why do we still have redundancies?

When we created the plots, we

were warned that

redundancies still exist in the

model. If we check the

degrees of freedom we see

that there are 11 redundant

constraints.

If you examine the mates, not

all concentric mates were

made flexible. For instance,

the mates between the piston,

cylinder and frame (that

were changed in the previous

exercise) are still hinge and

concentric mates in this

model. We do not have to

change the mates in this

problem because we are not

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Examine the other mates that have not been made flexible. These mate

concern forces or motion in the global Y direction (across the plane of

symmetry). As we expect these forces are going to be zero, we are not

concerned with these forces and do not have to take the time to remove

these redundancies.



Open Scissor_Lift from the Lesson08\Exercises\Redundancies Removal with Bushings\completed-optimum stiffness folder.


This is exactly the same assembly as used in the previous steps except

that the stiffness of the flexible mates has been changed.

10 Examine the bushings.

Edit one of the Concentric mates that is flexible.


Notice that Bushing has been selected and the

values set as follows:

Translational


� Stiffness = 100,000 N/mm

� Damping = 2000.0 N-s/mm

� Force = 0 N-mm

Torsional


� Stiffness = 100 N-mm/deg

� Damping = 20.0 N-mm-s/deg

� Toque = 0.0 N-mm

These values for stiffness and damping are more

realistic for a practical system then those used in

the previous example.

11 Run the study.

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The plots for the same four mates examined in the previous example

have already be generated.

As before, the plots for the symmetric pairs of mates are identical.

With the higher stiffness, we can see that after the initial acceleration,

the motion is sinusoidal.

Add the two values for the maximum force (ignoring the initial spike in

the magnitude) in mates Concentric20 and Concentric21, they

should be approximately 9,500 N which compares favorably with the

result obtain in Exercise 14: Kinematic Mechanism.

Also compare the values for the maximum force in mates

Concentric36 and Concentric37, they should be approximately

15,000 N which also compares favorably with the result obtain in

Exercise 14: Kinematic Mechanism.

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From these results, we can see that the forces were equal when we had

all the mates on one side of the model to the total force when we

removed the redundancies and split the force to the two sides.


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SolidWorks 2011 Exercise 18 Catapult

251

Exercise 18:Catapult

This exercise will further examine the

use of local flexible mates to properly

calculate the forces where multiple

supports are used.

We will use the same catapult model that

we saw in Lesson 3.

With a lot of redundancies, SolidWorks

Motion will solve the kinematics

correctly, however the force distribution

may be incorrect.


skills:





Project Description

Calculate the forces on the pivots between the arm and counterweight.


Open Catapult-assembly from Lesson08\Exercises\Catapult folder.

This assembly has been set up and run in the study named original study with results.


The counterweight is connected to

the arm with a single mate,

ConcentricB. There is also

Coincident4 which holds the

counterweight centered on the arm.

3 Play the simulation.

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Exercise 18 SolidWorks 2011 Catapult

252

4 Degrees of Freedom.

This assembly has 54

redundant constraints, but it

runs without problems. While

the kinematic problem is

solved, the problem with the

solution is that the resulting

forces may not be distributed

properly.

5 Create plot.

Create a plot of the global

Y direction resultant force

on the mate

ConcentricB.

We can observe a force of

about -1.22 N while the

arm is being rotated into

position, lifting the

counterweight.

Depending on our engineering judgement, this result may be good

enough if we are confident that the load is shared equally as we can just

divide the result in half to get the correct force at each pivot.

If this is not good enough, then we must make the mates flexible to

distribute the force correctly.

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6 Add another mate.

We are interested in the force

in each of the pivots between

the arm and counterweight,

so we need to have a mate on

the other pivot.

Select the Model tab, then

add a Concentric mate to the

other pivot.

Rename this mate

Concentric C.

7 Run.

Make sure that Replace redundant mate with bushings is cleared,

then rerun the study.

8 Create plot.

Create an additional plot showing the reaction force in the Y direction

(global coordinates) for the mate Concentric C.

We can see that the force is distributed evenly between the two mates.

This, however, may be just a coincidence as the distribution will

depends on how the software removes the redundancies. We will not

use flexible mates to ensure the correct force redistribution.

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9 Create local flexible mates.

Edit mates ConcentricB and ConcentricC.

Select the Analysis tab and select Bushing. Keep the default values.

10 Run.

Make sure that Replace redundant mates with bushings is cleared,

then, rerun the study.

We will still have a lot of redundancies in the model, however these do

not affect the results we are interested in at the two pivots.


The plots now show that the force is divided over the two mates.

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12 Change scale.

To make them easier to read, modify the two plots to show the Y Axis:

� Start Point = -2

� End Point = 1.0

� Major Units = 0.5.

We can see that the forces are exactly the same.


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Lesson 9Export to FEA


� Create an Action Only Moment.

� Export loads from SolidWorks Motion to FEA Simulation.

� Run the structural analysis in SolidWorks Simulation.

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Lesson 9 SolidWorks 2011Export to FEA

258

Exporting Results

Determining the forces on a part is generally not the end of the analysis

of a part. Usually, the forces obtained are to be used in finite element

analysis to determine the strength, displacement and Factor of Safety of

the individual parts. SolidWorks Motion and SolidWorks Simulation

work together to make the exporting of data from SolidWorks Motion

to SolidWorks Simulation seamless.

Case Study: Drive Shaft

The Drive Shaft assembly is composed of 5 sub-assemblies, and 2

single parts. SolidWorks Motion will be used to determine the forces

acting on one component, the Journal-cross and then using

SolidWorks Simulation, we will determine the stress and displacements

of the part.

Project Description

The universal joint is required to transmit a torque of 15,000,000 N-

mm at a speed of 2800 RPM. Determine the stress and deflection of the

part Journal_Cross_output.

Input_housing

Output_housing

Driveshaft

Output _housing

Output_shaft

Journal-cross_output

Input_housing

Input_shaft

Journal_cross_input

Driveshaft

Driveshaft

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SolidWorks 2011 Lesson 9Export to FEA

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� Create the motion study.

Using the known data as input, create a motions study.

� Run the motion study.

The motion study is calculated to determine the forces on the part

or parts in question.

� Export loads to Analysis.

From SolidWorks Motion, export the loads directly to SolidWorks

Simulation.

� Open the part for analysis.

Open the specific part in its own window.

� Run the FEA simulation.

Complete the boundary conditions in SolidWorks Simulation and

run the analysis.

� Examine the results.

Use the results to determine if design changes are needed.


Open Drive_shaft_assembly from the Lesson09\Case Studies

folder.

2 Insert a new motion study.

Make sure the units are in MMGS.

3 Add a motor.

Add a Rotary Motor to the Input_shaft. Make it turn at 16,800 deg/

sec (2,800 RPM).

Note the direction of rotation. It doesn’t matter which direction it turns

only that we know the direction so that we can add the Action Only

moment in the opposite direction in the next step.

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4 Add a force.

Apply an Action only Torque on the Output_shaft. This is a torque

that opposes the rotation, so set the direction opposite the motor added

in the previous step.

Input a value of 15,000,000 N-mm for the torque.

5 Study properties.

We are going to run the study for only 0.05 seconds, so we will need a

high frame rate to capture enough information.

Set the frame rate at 2,000. This will give us 101 frames.

6 Run.

Run the study for 0.05 seconds.

The following message will indicate that the current setting for the

Number of Frames parameter seem to be excessive and may

negatively impact the performance:

The playback speed or frames per second for this motion study will result in poor performance given the current study duration. Would you like these settings adjusted for better playback?

Click No to complete the simulation with the current settings.

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7 Calculate DOF.

We can see that there are zero

degrees of freedom, so we have a

kinematic system.

Close the Degrees of Freedom

window.

8 Examine the mates.

This assembly has zero DOF because of the way it was built. If you

examine the individual mates, many of them are point to point or point

to line to avoid removing too many degrees of freedom.

9 Plot results.

Create plots of the Angular Velocity Magnitude for both the input and

output shaft.

We can see that both shafts are turning at 16,800 deg/sec which was the

input speed.

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10 Plot the angular velocity of the driveshaft.

Create a plot of the Angular Velocity Magnitude of the driveshaft. We

can see the expected variation of the velocity caused by the offset angle

between the input and output.

11 Plot the required torque.

Create a plot of the torque of the input rotary motor. This is the torque

required by the motor to move the shaft at this load.

FEA Export Motion Simulation enables you to apply all of the necessary resulting

quantities (forces, moments, accelerations etc.) onto the load bearing

faces and solve for the for the stress and deformation analysis

(SolidWorks Simulation module is required for the deformation

solution). This way, Motion Simulation simplifies the transient problem

using the rigid body dynamics approach and solves for the parts’

accelerations and joint reaction forces. Then, in SolidWorks

Simulation, these loads are applied on the bearing faces and the stress

analysis problem is solved.

Motion simulation enables you to apply the loads and solve the

deformation analysis using SolidWorks Simulation in two distinct

ways:

� Direct solution, where the setup, solution and the post-processing

is performed directly in the Motion Simulation interface.

� Export of the loads into SolidWorks Simulation. Deformation

solution is performed using SolidWorks Simulation interface.

Both approaches are presented in this lesson.

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Load Bearing Faces

The applied (or exported) forces are

transferred onto the faces only; edges

and points are not allowed. Any face

used in the mate definition in

SolidWorks is also assumed to be the

load bearing area for the applied (or

exported) loads. If other entity types

(points, edges) are used in the mate

definition, load bearing faces have to be

specified under the Analysis tab.

Mate location Default initial location of the mate in motion analysis is determined

using the first entity in the definition of the mate. For example, in the

mate definition shown in the figure above, the initial mate location is at

the center of Face<1>@Input_shaft-1/universal_bearing-1.

Optionally, this can be changed by selecting a new entity in the Mate

location field. Changing the location of the mate may change the

motion analysis results and the resulting reaction forces somewhat; the

impact of this change varies from case to case.

It is recommended that you change the mate location if the initial

configuration is not suitable. This can be especially important when

using the motion loads for the finite element analysis using the

SolidWorks Simulation modulus.

Motion Simulation also exports the body loads due to the accelerations

of the parts. Similarly to the joint reaction forces, body loads are

exported at each (or all) requested time step.

The load bearing faces as well as the new mate locations have to be

input before running the motion analysis.

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Export of Loads This section of the lesson demonstrates how to properly prepare a part

subjected to the motion loads for the finite element analysis in

SolidWorks Simulation. First, the correct load bearing faces and mate

locations will be defined. Then, the motion loads are imported in the

SolidWorks Simulation, where the finite element analysis and the post-

processing are performed.

12 Isolate on journal_cross<1>.

This is the journal_cross on the input

side of the driveshaft. Isolating this

component is done just to make it easier to

see the part.

We are interested in computing the

stresses and displacements of this part.

Examine the four mates of this part. None

of the mating entities are faces, but rather

points or axes. This will require us to

specify the faces where the forces will be

transferred for each of these four mates.

Click Exit Isolate.

13 Specify the load bearing faces.

Edit the first mate, Coincident24.


Select Load Bearing Faces.

Click Isolate components. This will hide

all components except those associated

with this mate.

Use the Select Other command to select

the exterior surface of the journal_cross

and the internal cylindrical face of the universal_bearing. The two

parts are shown in the exploded view below for clarity.

Because the faces are touching, Treat as Bonded if touching is

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As mentioned in the previous discussion block, the default initial mate

location is determined from the first entity in the mate definition, center

of the Face<1>@Input_shaft-1/universal_bearing-1. Because

these two parts are in permanent contact and do not translate

significantly relative to each other, the default location of the mate at

the center of the above face is acceptable and would not have to be

changed. It is, nevertheless, a good habit to place the initial location to

the most ideal location, especially if we intend to follow with the finite

element stress analysis of a part. To practice this we will change the

location for all four journal_cross-1 mates.

Selecting the mate location is optional. You can select either of the

points that define the mate, but it is not necessary.

Click OK and Exit Isolate.

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14 Second load bearing face.

Edit mate Coincident25.


Select Load Bearing Faces.

Click Isolate components.

Select the two faces shown, one on the journal_cross and the other on

the attachment flange.

As the faces are not touching, the option to Treat as Bonded if

touching is not available.

15 Define remaining load bearing faces.

Repeat the above procedure on the remaining two mates,

Coincident26 and Coincident28.

16 Re-run the analysis and Save the assembly.

After the mate locations have been changed, the motion analysis must

be re-calculated.

We will now proceed with a stress analysis of the journal_cross-1

component.

SolidWorks Simulation Users Only

SolidWorks Simulation may read the Motion loads for a single time

step or a multiple time steps at once. In the latter case a design scenario

feature of the Simulation software is used to run multiple analyses at all

requested time steps. Design Study enables us to locate the critical time

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17 Import motion loads.

Ensure that SolidWorks

Simulation is added inside

SolidWorks.

Click Import Motion

Loads on the Simulation

menu.

Select the Motion Study

from the list that you used

to create the forces.

Select journal_cross-1 in Available assembly

components, then click >

to move it to the Selected

components box.

Click Multiple frame

study.

In the Start (Frame No.) box, type 80.

The End box should already be 101.

Click OK. This will import and save the load data to the CWR file for

the journal_cross-1 part, and define the design study.

The above specifications define design study with 22 sets. Each set has

its loads defined from the motion loads developed at the time instant of

the frame associated with that set.

18 Open the part.

Open the part journal_cross-1 in its own window.

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19 Select the SolidWorks Simulation study.

A new static study named CM-ALT-Frames-80-101-1 has been

added. The numbers 80, 101 and 1 in the study name refer to the

starting and ending frame numbers and the frame increment,

respectively.

20 Select the Design Study.

A new design study named CM-ALT-Frames-80-101-1 has also been

added. You can review the list of the parameters along with their values

that have been imported from SolidWorks Motion. 22 scenarios

corresponding to the frames 80 to 101 have been created.

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21 Apply material.

The definition of the static study needs to

be completed.

Back in the static study, apply Alloy Steel to the part.

In the Simulation Study tree, right-click

the journal_cross part and click Apply/

Edit material.

Select Alloy Steel from the SolidWorks Materials library files.

Click OK.

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22 Mesh the part.

Right-click Mesh in the Simulation Study tree

and click Create Mesh.

Under Advanced, verify that Draft Quality

Mesh is cleared.

Move the Mesh Density slider to set the

Maximum Element Size close to the value of 30

mm.

Click OK and the model will mesh.


Right-click the study

icon and click

Properties.

Because this part is self

equilibrated, Use

inertial relief is on by

default.

Click OK.

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Note The inertial relief is one of the options used to stabilize self-

equilibrated problems in the finite element analysis. The detailed

discussion of the option is a subject of the SolidWorks Simulation

course.

24 Run Design Study.

Select the design study tab and click the Run button.

The 22 different sets of data will be solved sequentially.

25 Global maximum for von Mises stress.

Global maximums indicate the maximum

values over all 22 scenarios.

In the design study tree, right-click Results and Graphs and select

Define Design History Graph.

Click Constrains for the Y-Axis and select VON:

von Mises Stress.

Click OK.

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The graph shows the variation of the maximum von Mises stress in the

journal_cross-1 part across all 22 scenarios. We can observe that the

largest value of 5.07 e8 N/m2 (507 MPa), reached in scenarios 1 and 22,

is smaller than the yield strength of the material (620.4 MPa).

27 Global maximum for resultant displacement.

Create a similar graph showing the global maximum of the resultant

displacements.

The maximum displacement is 0.12 mm.

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28 Von Mises Stress plot for design scenario #15.

Design study stores full results for all

computed scenarios.

In the design study, click the column

corresponding to Scenario 15 to access the

results.

Under Results and Graphs, double-click the

VON: von Mises Stress plot.

The maximum von Mises stress magnitude in scenario #15 is 491 MPa.

29 Resultant Displacement plot for design scenario #15.

The maximum resultant displacement in scenario 15 is 0.12 mm.

30 Save and close the journal_cross part file.PRE-

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Direct Solution in SolidWorks Motion

This section of the lesson demonstrates how to perform stress analysis

on a part directly in the SolidWorks Motion interface.

Important! The correct load bearing faces and the mate locations specified in steps

13 to 15 must be specified for the direct stress solution in the Motion

Simulation as well.

Note SolidWorks Simulation modulus must be activated in order to obtain

the stress solution.r

31 Simulation setup.

In the Drive_Shaft_Assembly motion study,

click the Simulation Setup icon .

In the Part for Simulation field select the

journal_cross<1> on the input side of the

driveshaft.

Specify 0.0395s and 0.05s for the Simulation

Start Time and Simulation End Time,

respectively. Click Add Time to add the time

range to the Simulation Time Steps and

Time Ranges field.

Under Advanced, move the mesh slider to set

the Mesh Density Scale Factor to 0.95 to

generate finer mesh.

Click OK.

The following message will display:

Do you want to assign material to the part?

Click Yes to open the Material window.

32 Material.

Similarly to step 21, specify Alloy Steel.

Click Apply and Close.

33 Solve finite element simulation.

Click the Calculate Simulation Results button .

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34 Stress results at 0.045s.

To show the result plot, move the time line to 0.045s.

Note The specified time must fall within the time range requested in step 31.

Set the Results Plot button to show the von

Mises Stress Plot.

The legend indicates the maximum stress of approximately 542 MPa.

However, because journal_cross<1> is shown in the context of the

whole assembly the stress contours are not easily visible. In this case

clearer plot will be provided when we isolate the part.

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35 Isolate on journal_cross<1>.

The stress contours are now visible. The indicated maximum of 542

MPa is below the yield strength of the material, 620.4 MPa.

36 Factor of Safety at 0.045s.

Follow steps 34 to 35 and show the plot of Factor of safety.

The minimum factor of safety indicated in the plot is 1.14 (620.4/

542=1.62).

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37 Deformation results at 0.045s.

The maximum resultant displacement at time 0.045s is 0.37 mm.

38 Results at different times.

Move the time line to any other time step. The contour plots will update

automatically.

Note The specified time must again fall within the time range requested in

step 31.

39 Animate and show the overall maximums.

To set the legend to show the overall maximum for the requested

analysis time range and to see the animation, click the Play button.

The maximum resultant displacement over the entire requested analysis

interval (0.035s - 0.05s) is 0.38 mm.

40 Save and close the file.PRE-

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Summary This lesson showed the procedure for the application of the joint and

body loads computed in the Motion Simulation in the finite element

stress analysis. In the first part we solved the rigid body dynamics

problem and obtained the necessary joint and body loads. Then, load

bearing faces and mate initial locations were specified in the assembly

mate definitions.

In the second part the loads from multiple time steps were applied on

the selected part and the stress analysis was carried out. Two

procedures are currently available: direct stress solution in the Motion

Simulation interface, or the export of the motion loads in the

SolidWorks Simulation. In the later case, the stress solution is carried

out in the SolidWorks Simulation interface with the help of the design

study feature.

The above procedures allowed us to locate the extreme stress in the part

of the rotating drive shaft assembly. Displacements, factor of safety and

other results available in the SolidWorks Simulation are available and

were shown in this lesson.

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SolidWorks 2011 Exercise 19Export to FEA

279

Exercise 19:Export to FEA

In this exercise, we will export

the loads for a latch mechanism

to SolidWorks Simulation and

conduct an analysis of the part.

This exercise reinforces the

following skills:

� Exporting Results on

page 258.

Project Description

Determine the maximum stress and deflection on the part J_Spring.


Open Full latch mechanism from the Lesson09\Exercises folder.

This is the same assembly used in Lesson 4.

The motion study is already set up and has been run.

2 Play the study.

Click Play (do not run) just to refresh your memory about how the

mechanism works.

3 Specify load bearing faces.

Locate mate Concentric6. This is the mate that is used as the pivot for

the spring.

Edit the mate and specify the four faces shown as the load bearing

faces. The two parts are shown in exploded view for clarity.

For the Mate location, select the edge of the split surface on either the

clip or pin.

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Exercise 19 SolidWorks 2011Export to FEA

280

4 Re-run the simulation.

Because the contact faces as well as the mate locations were changed,

the motion simulation needs to be recalculated.

Contact Forces While the forces in the mates can be imported to SolidWorks

Simulation automatically, the contact forces cannot and must be

defined manually.

We will first determine where the contact forces are maximum through

observing the plots created in SolidWorks Motion. We will then

determine the frame at which this maximum force occurs so that we

only have to output the data for a single frame.

We must also determine the directions along which these forces must

be applied.

5 Examine the plot of contact force.

The plot of the magnitude contact force between the J-spring and the

keeper is already created. We can see that the maximum force occurs

at about 2.4 seconds.

6 Create additional plots.

Plot the contact forces for both the X (red) and Z

(blue) components. We expect the Y (green)

component to be zero, or nearly so, therefore it is

not important to the analysis.

Note By default, the forces are output in the assembly global coordinate

system.

Fmagnitude, max

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Note You do not have to select the actual contact faces, only the components.

Ignoring the short duration peaks, notice that at the point where the

X component is maximum, the overall magnitude is not at its

maximum. We will run the analysis at the point where the overall

magnitude is maximum.

For additional practice, run the analysis at the point where the

X component is maximum.

Fx, max

Fz, max

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7 Create an additional plot.

Create a plot of the joint reaction force for the mate linking the

J_spring to the knurled_pin, Concentric6.

Modify the Y axis of the plot so that the End Point is 50.

Compare this plot to the first plot of the contact force. Both plots

should be exactly the same, i.e. the magnitude of this force must be the

same as the contact force.

Close the plot.

8 Modify the plot to show frames.

Modify the contact force magnitude plot to show

the X axis in Frames.

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9 Change the axis range.

To better see the area of interest, modify X axis of the plot to show

from frame 320 to frame 340.

To make the graph easier to read, change the X axis major and minor

units to 10 and 5 respectively.

We can see that the most extreme loading occurs at about frame 325.

When we export the motion loads to SolidWorks Simulation, we will

export the data from just this one frame.

10 Modify plots.

Change the X axis to Frames for both the X and Z contact force plots.

11 Export forces.

In step 6 we determined that the

two directions of interest were X

and Z as shown in the image. We

will just export these as the Y

direction should essentially be

zero.

Export both the X and Z contact

forces for the J_spring to CSV

files. Right-click each plot and

click Export CSV.

Each file will get a default name and be saved in the same directory as

the assembly.

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12 Examine the output data.

Open each of the two CSV files and note the values at frame 325.

13 Export Motion Loads.

When the calculation

completes, save the result

and export the loads for the

J_spring for the frame

325 only.

14 Open the part.

Open the J_spring part in

its own window.

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15 Simulation study.

Select the simulation

tab for the new study CM1-ALT-Frame-325.

The mate loads have

been imported into the

part, but we will have

to apply the contact

forces manually.

Notice the directions of the global coordinate system for this part are

different from the assembly. The X direction in the assembly is the

Y direction in the part and the Z direction in the assembly is the

X direction in the part. When we apply the contact forces to this part,

we will have to insure we are using the correct force for the direction

on the part.

16 Apply X contact force.

Apply a force of -34.80615053 (from the exported values in the CSV

file) to the indicated face. Select the Right plane to define the

direction. As the negative value of the force was the reaction force, we

have to reverse its direction to be correct in the part.

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17 Apply Z contact force.

Apply a force of 9.569044324 to the indicated edge. Select the Top

plane to define the direction. Reverse its direction to be correct in the

part.

18 Apply material.

Apply the material Alloy Steel to the part in the Simulation Study tree.

19 Mesh the model.

Right-click Mesh in the

Simulation Study tree and

click Create Mesh.

Use the default settings.

Click OK.

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20 Run the study.

Right-click the study and click Run.

We will get a warning that says:

Warning: There is a significant external imbalance force in the X-direction which will be balanced by the application of opposing inertia forces. Unless you model is under such a force or under marginally imbalance forces, application of Inertia Relief may alter the characteristics of your model.

This message is the result of exporting the loads from the motion

simulation and entering values by hand. This part can therefore be

considered as nearly self equilibrated. Click Yes.

Click Yes.

21 Stress Plot.

Examine the stress plot. We can see that the maximum stress 150 MPa

and is on the underside of the J_spring.

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22 Factor of Safety.

Create a Factor of Safety plot to determine if the part is yielding.

Right-click the Results folder and click Define Factor of Safety Plot.

Use the default values to create a plot that shows the Factor of safety

distribution.

Click OK.


We can see that the minimum Factor of Safety is 4.12, so the part is not

yielding.


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Lesson 10Event Based Simulation


� Understand and run event based simulation.

� Apply servo motors.

� Create events with specific timing and logic.

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Lesson 10 SolidWorks 2011Event Based Simulation

290

Event Based Simulation

This lesson introduces the event based motion simulation of the

mechanism, which incorporates the event-triggered control.

Case Study: Sorting Device

The sorting device shown in the

figure is used to sort two types of

boxes: yellow with the hole and the

solid brown. Each type should be

moved to the corresponding bay.

Event based simulation will be used

to simulate this mechanism.

Problem Description

The mechanism used to sort the boxes into the respective bays consists

of six parts. The vertical motion of the boxes is caused by the gravity.

The horizontal motions are then driven by a set of three pistons with

servo motors. Motors actuate the motion based on a set of sensors

controlling the box type and their position in the mechanism.

Simulate a mechanism placing each box type into its respective bay.


Open Sorting device from the Lesson10\Case Studies folder.

2 Verify the units.



Name the study Sorting device.

Servo motors Servo motors are both rotational and linear motor features driving

mechanisms in event based simulation. Their motion is, however, not

prescribed directly in the Motor FeatureManager. It is controlled via an

event based simulation interface, and it can be triggered based on

various criteria such as proximity of a certain part in the system.

Introducing: Servo

Motors

Servo motors are used as motion drivers in the event based simulations.

Where to Find It � On the MotionManager toolbar, click Motor . Under Motion

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SolidWorks 2011 Lesson 10Event Based Simulation

291

4 Servo motor #1.

Define linear Servo Motor for Actuator<1>.

Click the Motor icon and select Linear Motor (Actuator).

Select the indicated face for both Motor Location and Motor

Direction.

Under Motion select Servo Motor and Displacement.

Rename this simulation component to Actuator 1.

Click OK.

5 Servo motor #2 and #3.

Similarly, define two more linear, displacement based servo motors for

Actuator<2> and Actuator<3>.

Rename the two motors to Actuator 2 and Actuator 3,

correspondingly.

Click OK.

Sensors Sensors can be used to trigger events or stop them. Three different

sensor types can be used in event based simulations:

� Interference detection sensor for detecting collisions.

� Proximity sensor, which detects motion of a body crossing a line.

� Dimension sensor used to detect the relative position of component

from dimensions.

Introducing: Sensor

s

Sensors can be used to trigger or stop the motion in event based

simulation.

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Where to Find It � In the SolidWorks FeatureManager, right-click Sensors and select

Add Sensor.

� On the Evaluate tab click the Sensor button .

6 Proximity sensor #1.

Two proximity sensors are used to control

the system. Sensor 1 is used to detect the

solid box on the bottom platform of the

holder. Sensor 2 is then tracking the

hollow box.

Define Proximity sensor

detecting the presence of the

solid box on the platform.

Select the indicated face of

sensor 1 for Proximity

sensor location. The

Proximity sensor direction

field may remain empty to

keep the default vertical

direction.

Select the two solid boxes for

the Components to track

field.

Enter 12 mm for Proximity

sensor range.

Click OK.

Rename this sensor to Sensor 1.

Note A 12 mm range was used to trigger the necessary

event when the box reaches the horizontal

platform of the holder. Since the thickness of the

platform is 10 mm, any sensor ray longer than 10

mm will trigger an event as the box approaches

the platform.

7 Proximity sensor #2.

Similarly, define proximity sensor #2 to track the boxes with the hole.

Rename this sensor to Sensor 2.

Sensor 1Sensor 2

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8 Contacts.

Define the following four solid bodies contacts.

Whenever possible, use contact groups to simplify the contact solution.

9 Gravity.

Define Gravity in the negative Y direction.

Task Event based simulation requires a set of tasks, triggered by sensors, and

ordered sequentially or overlapping in time. Each task is defined by a

triggering event and its associated task action, which controls or

defines motion during the task.

Triggers Each task is triggered with a triggering condition. The triggering

condition can depend on the status of a sensor, or start or end of some

other task in the sequence.

Contact group 2(Steel (Dry))

Contact group 1(Acrylic)





All in one group(Acrylic)

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Task Action The following is a list of actions which can be specified in a task

definition.

� Stop . Stop the motion of a component.

� Motors . Turn on or off any motor, or change a constant speed of

motor according to the selected profile.

� Forces . Apply or stop applying any force, or change a constant

force according to a selected profile.

� Mates . Toggle the suppression of a selected mate.

Timeline View vs.

Event-based Motion

View

To define the task, Motion Simulation offers an Event-based Motion

View which can be accessed through the corresponding button on the

MotionManager toolbar . This view is used to define tasks and

design the logic of the system.

Timeline based view provides classical Motion simulation view with

the keys, indicating the beginning, end and change in the behavior of

the simulation components. The sequence of the keys in time is

generated when the event based simulation computations complete and

is also an important result of the simulation.

Tasks design tableTasks time sequenceand logical relationship

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Introducing: Task Task controls and defines motion of the components during the

simulation. It is defined by a triggering event and its associated action.

Where to Find It � On the MotionManager toolbar, click the Event-based Motion

View button. To add a task, click the Click here to add line on

the bottom of the task list table.

10 Event-based motion view.

Switch to the Event-based Motion View.

11 Task #1 - Name and Triggers.

The first task for the system is to move the lowest solid box along the

holder platform to the position, where Actuator 2 may push it into

Bay 1. This task will be triggered when the bottom solid box activates

the proximity Sensor 1. Because this sensor triggers an event when

the solid box is 2 mm above the platform, and to provide enough time

for Actuator 2 to fully retract, a 0.1s time delay for this task will be

specified.

Click Click here to add line to add a new task line.

Enter Push solid box for Name.

Click the selection button in the Trigger

field to open the Trigger dialog window.

Select Sensor 1.

Click OK to close the Trigger dialog

window.

Back in the Event-based Motion View,

complete the Triggers section by setting Condition to Alert On and

Time/Delay to 0.1s (see the figure in the next step).

Click here to adda new task line

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12 Task #1 - Action.

The task definition will complete with the specification of action. In

this case the action comprises of Actuator 1 motor pushing the solid

box 75 mm along the platform. (this is an ideal position for subsequent

action of Actuator 2).

Select Actuator 1 from the Motors in the Feature field, Change for

Action, 75mm for Value, 1s for Duration and choose Harmonic for

Profile.

13 Task #2 - retracting Actuator 1.

Define the second task to retract Actuator 1. This task should be

triggered after the task #1, Push solid box, completes and its duration

is 0.2s.

Name this task Retract Actuator 1.

14 Task #3 - pushing solid box into Bay 1.

Define the third task, this time for Actuator 2, pushing the solid box

into Bay 1. This task comprises of a 50mm extension of Actuator 2

in 0.6s.

Similarly to the task #2, Retract Actuator 1, the task #3 is triggered

by the completion of the task #1, Push solid box.

Rename this task to Push solid box to Bay 1.

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15 Task #4 - retracting Actuator 2.

Similarly to step 13, define the task #4 retracting Actuator 2 in 0.1s.

This task is triggered by the completion of task Push solid box to Bay 1.

Rename this task Retract Actuator 2.

16 Tasks for boxes with hole.

Follow steps 11 to 15 and specify similar tasks to move the box with

the hole into Bay 2.

To move the box with hole next to Actuator 3, extend Actuator 1 by

130 mm in 1.2s with a 0.1s delay. Then retract Actuator 1 in 0.3s.

Use the same time and distance values for Actuator 3 as those which

were used for Actuator 2 in steps 14 and 15.

Give the new tasks names similar to those used in steps 11 to 15.

17 Simulation Properties.

Set the Frames per second to 200 and check Use Precise Contact.

Under Advanced Options, set Maximum Integrator Step Size to

0.05s.

Note To speed up the simulation and because we are not interested in the

force results, the maximum integrator step size value can be relaxed.

18 Calculate simulation for 7 seconds.

It takes approximately 15 minutes for the simulation to complete.

19 Animate.

Animate the final motion of the system.PRE-

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20 Timeline View.

Switch to the Timeline View. Here you can see the result of the event

based simulation. Each key indicates the beginning, end or a change in

the motion of the system components. It also indicates the duration of

the whole cycle.

The Timeline View provides insight into the duration of the whole

operation. Each task start and end is identified with a time key. Possible

action following this simulation may be a change in the velocities of

the actuators in order to optimize the system, change of material in

order to change the effect of friction, change of the design to better

stack the boxes in the bays, or similar.


Summary In this lesson, we introduced the event based motion simulation to

model and optimize the behavior of mechanical systems. We modeled

an operation of the sorting device mechanism used to move two

different box types into their respective bays.

The event based simulation is a sequence of tasks triggered by sensors,

ordered sequentially or overlapping in time. Each task is defined by a

triggering event and its associated task action which controls or defines

motion during the task. The triggering condition can depend on the

status of a sensor, or the start or end of some other task in the sequence.

The triggered action can be applied to motors, forces, mates or can

completely stop the motion of a moving part. A special type of motor,

servo motor, was introduces and utilized.

The tasks with their respective actions, sequence and the logic is

designed in the Event-based Motion View interface introduced in this

lesson.

The result of the simulation, the animation and the duration as well as

time sequencing of the entire operation was shown. This result can then

be used to modify the system parameters (kinematic parameters of the

actuators, for example) to optimize it.

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Lesson 11Design Project (Optional)


� Create a function based force.

� Export loads to SolidWorks Simulation.

� Complete an analysis project from motion to FEA.

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Lesson 11 SolidWorks 2011Design Project (Optional)

300

Design Project This lesson is in two parts, in the first section it is up to the individual

students to solve the problem of the Surgical Shear. In the second part

of the lesson, the complete solution will be shown.

The Case Study is also in two parts:

The overall problem is to determine the suitability of the design of the

handle used in the surgical shear.

Part 1: Develop loads on the parts based on a motion study. This will

require the development of a force function to simulate the resistance

of the catheter being cut by the surgical shear.

Part 2: Conduct an FEA simulation of the handle, using the loads

developed in part 1.

Case Study: Surgical Shear - Part 1

The Surgical Shear is used to cut

arteries and catheters. It consists of

a fixed blade and a moving blade.

Because the surgical shear has to be

operated by many people in the

medical industry, it is important to

estimate the handle force that will

be sufficient to generate the required cutting force.

In this part of the lesson, we will mate the components, create a motion

study and develop a force function to simulate the blades cutting

through a catheter.

Problem Description

The mechanism is composed of seven parts. The fixed_cutter is

stationary and the rotation of the handle provides the motion of the

moving_cutter. The latch rotates with respect to the fixed_cutter

and is inserted into the moving_cutter. A spring maintains the

moving_cutter in an open configuration when no force is applied to

the handle. The removable blades are attached to the fixed_cutter

and moving_cutter parts.

When the surgeon squeezes the handle, it will rotate 12 degrees and

move the blade. The spring is used to help to return the shear to the

open position. It is assumed that it takes a surgeon approximately one

second to cut the catheter.

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SolidWorks 2011 Lesson 11Design Project (Optional)

301

Determine the design suitability of the handle part.

Force to Cut the Catheter

From experimentation, the force to cut a 3 mm catheter was determined

and plotted in the graph below. The X axis shows the travel of the blade

starting at Point 1(X=0mm) where the blade contacts the catheter. The

cutting force increases slowly at first as the catheter is compressed and

then climbs more rapidly as we approach the point when the cutting

begins.

At Point 2 (X=1.5mm) the blade starts cutting the catheter and the

force reduces quickly as the cut portion of the catheter returns to its

round shape.

From Point 3 to Point 4, the remaining thickness is cut and at Point 4

the cut is complete.

blade1

blade2

moving_cutter

handle

latch handle_link

fixed_cutter

Point 1

Point 2

Point 3

Point 4

X

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The preceding graph contains the variation of the cutting force from the

experiment. In order to input this curve in SolidWorks Motion

Simulation each segment has to be expressed as function of the catheter

location (location of the cutting blade); in the above graph this location

is expressed through the variable “x”. Notice that it assumes values

from 0 mm (cutting blade touches the catheter) to 3mm (cutting blade

completes the cut). Each segment is therefore expressed as a linear

function shown in the graph.

Note The above cutting force curve as measured from the experiment is

expressed as function of the blade location (not the function of time).

Time dependent data is not available since, in general, it depends on

how fast the cutting operation is completed and how the input force

from the surgeon’s hand varies in time. While the input of the time

dependent force would be trivial (this procedure was practiced multiple

times throughout this course), input of the location dependent force is

more challenging.

Note also, that with certain assumption the above location dependent

function can be converted in time dependent input. However, to

demonstrate the more complex case (which may be required in some

analyses) we will use the location driven input.

As we will see, inputting the number of data segments into the force

expression can be tedious. We will simplify this curve as shown in red,

with only three segments; this should be enough to reasonably simulate

the cutting force.

First Segment

Second Segment

Third Segment

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Each of the three segments can be defined by a linear equation:

First Segment: y = 7.333333 x

Second Segment: y = -80 x + 131

Third Segment: y = -2.14286 x + 6.42857

Self Guided Problem - Part 1

In this section, it is up to each student to solve this motion part of the

problem. A basic outline of the procedure is provided as a guide, but

the details of the steps are left up to the student.


The basic steps are:

� Add mates.

Add the appropriate mates to insure that the mechanism operates as

specified.

� Determine the cutting force.

The action/reaction force on the blades while cutting a catheter has

been experimentally determined and is not linear. From the

experimental data, develop an expression to simulate this force on

the blades.

� Run the motion analysis.

Run the study and create the appropriate plots.

� Analyze the mechanics.

Interference detection is run to make sure that the mechanism will

move through its full range of motion. Load paths are examined to

ensure that the correct forces have been calculated in the

simulation.

Note You can view the Surgical_shear.avi movie to help you understand

the mechanism motion.

The steps below outline the procedure to form a road map of the

necessary steps:


Open Surgical_shear from the Lesson11\Case Studies folder.

When opened, the components are not mated.

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2 Mate the components.

It is up to the student to

determine the best

method to mate the

components to reflect the

mechanical operation of

this mechanism and

reduce redundancies.

Keep in mind that the part

of interest in this project

is the handle.

3 Add motion drivers.

Add appropriate motors and springs to capture the design motion (see

the problem description).

4 Develop a distance based force.

The action/reaction force developed by cutting the catheter is not

linear. An expression for this force, based on the position of the blades,

must be developed to simulate the experimentally determined forces.

5 Analyze the results.

Create plots and check for interference.

Modify parts as necessary.

Self Guided Problem - Part 2

In this section, it is up to each student to solve this FEA part of the

problem. A basic outline of the procedure is provided as a guide, but

the details of the steps are left up to the student.


� Export loads to SolidWorks Simulation.

Once we have the loads calculated, they are exported to SolidWorks

Simulation to evaluate the suitability of the parts.

� Replace motion drivers.

Some motion drivers such as motors need to be replaced with

forces or moments in order to run a static analysis.

� Analyze part.

Analyze the part using SolidWorks Simulation to determine the

suitability of the part based on strength and deflection.

� Refine part.

If the analysis determines that the part is not suitable as designed,

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Problem Solution - Part 1

In this section, the complete solution to this problem is provided.


Open Surgical_shear from the Lesson11\Case Studies folder.

Note Mate names are normally not important, however to ensure that mates

described in the text are consistent with the model, specific mate names

are given in the following steps. If you apply mates in a different order,

just rename the mates to be consistent with the images.

2 Lock mates.

The two blades are rigidly connected to the fixed

and moving cutters, so the appropriate mate would

be the lock mate.

3 Coincident mates.

The moving_cutter slides along the

outside faces of the fixed_cutter.

Two Coincident mates, between the

faces shown and the corresponding

faces on the moving_cutter, can be

used to hold this relationship.

4 Mating the linkage.

We need three mates to connect the

linkage.

A Hinge mate can be used to connect

the handle to the fixed_cutter, and

another Hinge mate to connect the

handle_link to the moving_cutter.

The remaining mate between the

handle and handle_link should be a

Concentric mate, preferably using

faces, to avoid over defining the mates.

Lock1

Lock2

fixed_cutter moving_cutter

Coincident1Coincident2

Hinge1

Concentric1

Hinge2

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5 Mating the latch mechanism.

The latch mechanism needs two different mates.

A Hinge mate is used to control the rotation and

position with respect to the fixed_cutter. The

selected surfaces are shown in the image (the

moving_cutter has been hidden).

A Cam mate could be used to mate the boss on the

latch to the slot in the moving_cutter. In our

initial solution to this problem, we will not use a

Cam mate, but will use Contact in the motion

study.

6 Set the initial position.

Before creating the motion study, we need the

blades to be 7.25 mm apart as the initial position.

Use a Position Only mate to set the distance.

Now that everything else is positioned, we

need to also make sure that the boss on the

latch is touching the appropriate surface.

Later, when we add contact and a spring,

those conditions will force the boss onto the

surface. We want to make sure however,

that when the motion study starts, the boss

doesn’t have to move onto the surface and

create a transient condition that would not

occur in the physical part.

Temporarily Fix the moving_cutter and use a Tangent Position Only

mate to set the boss on the latch against the surface of the slot. When

finished, Float the moving_cutter.

7 Create a new motion study.

Cam

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8 Add a spring.

A linear spring is used to

connect the latch to

fixed_cutter.

The spring has a stiffness of

0.175 N/mm and free length of

40 mm.

9 Add Contact.

Add Solid Bodies contact between the latch and the moving_cutter.

Specify Steel (Dry) for the material.

Select Friction, both kinematic and static.

Click OK.

10 Add a rotary motor.

When the surgeon uses the shear,

the squeezing of the handle will

rotate it 12 degrees. It takes

approximately 1 second to

squeeze and open the shear. To

simulate this action, we will add a

Rotary Motor.

The motor parameters should be

Oscillating, 12deg at 1 Hz with

0deg Phase Shift.


Set the properties to record 100 Frames per second.

Select Use precise contact.

Verify that Replace redundant mates with bushings is cleared.

12 Run the simulation for 1 second.PRE-

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Creating the Force Function

Now, we need to create an action/

reaction force that will simulate the

resistance effect of cutting the catheter.

We will consider a catheter with 3 mm

diameter.

The force will have to be defined in

several steps based on the physical

conditions. Before creating an

expression, we should be able to

describe the motion in words:

Step 1: During the initial movement of the blade, there is no force as

the blade is moving through open air.

Step 2: Once the blade contacts the catheter, there is a resistance as the

catheter is compressed before it is actually cut.

Step 3: The catheter is cut and the force is rapidly reduced.

Step 4: The catheter is cut, but the blade continues forward without

resistance.

Step 5: The blade moves back to the starting position without

resistance.

Steps 1, 4 and 5 above are easy as they are just zero, while our real

problem is defining the force in Steps 2 and 3.

Force to Cut the Catheter

The experimental data was shown on page 301. The two graphs are

repeated

below.

Point 1

Point 2

Point 3

Point 4

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The simplified Cutting Force plots is shown again with the equations

for the three segments.

Each of the three segments can be defined by a linear equation:

First Segment: y = 7.333333 x

Second Segment: y = -80 x + 131

Third Segment: y = -2.14286 x + 6.42857

Stages in the

Process

We will build the full expression piece by piece to see how it is

constructed.

� Create a variable for the location of the cutting blade as it cuts

through the catheter (variable x in the above graph).

� Plot the force function for the first segment

� Plot the force function for the first and second segments

� Plot the force function for the first, second and third segments

� Terminate the force function when the cutting blade cuts through

the catheter completely (x=3mm)

� Set the force function to zero in the second part of the cutting

process when the cutting blade moves in the opposite direction.

First Segment

Second Segment

Third Segment

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13 Determine the blade clearance.

Measure distance between blades. The distance is 7.25 mm, so if the

catheter is 3mm, then at the start there is a clearance of 4.25 mm.

14 Create plot of the displacement between

blades.

Select the two vertices in the order shown. If you

select them in the reverse order, the plot will be

reversed.

As the force is a function of blade position, we

need to know the position of the blades. By

creating the plot, we have a variable to use in the

expression.

Important! The remainder of this section of the lesson assumes that this is the first

linear displacement plot and therefore its name is

LinearDisplacement1. Likewise the force we are about to add will

be Force1 and the linear velocity plot created later will be

LinearVelocity1. If you have created other plots or forces and the plot

names you obtain in these steps are different, you must either rename

your plots or substitute your plot names as appropriate.

1 2

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Creating the Force Expression

The goal in the next section is to develop an

action/re-action force between the two blades

that represents the force necessary to cut the

catheter.

Rather than apply the force directly to the blades

while developing the force expression, we will

use a dummy force that will not affect the

outcome of the motion analysis. We are going to apply this force to the

fixed_cutter. As SolidWorks Motion is a rigid body analysis tool, any

force applied to a fixed part can have no effect on the motion analysis.

15 Add a force.

This force does not affect the results as it is

applied to a non-moving part. We will use it to

develop the full expression for action/reaction

force.

Define this force as a function equal to the

Linear Displacement1 defined in the previous

plot.

16 Run.

Run the simulation for 1 second.

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17 Create a plot.

Plot the Y Component for the Reaction Force. The force is now

directly related to the position of the moving blade.

w


The above force starts at -7.25 because the blades are 7.25 mm apart at

the beginning of the simulation. The force is zero when the distance

between the blades is zero. In our simulation however, we want the

force to be zero when the blades are 3 mm apart (when the blade first

contacts the catheter).

Change the force expression to be {LinearDisplacement1} + 3.


The force now starts at -4.25 because that is the distance from the blade

to the catheter. The force is now zero at contact between the blade and

catheter.

Important! This last expression, {LinearDisplacement1} + 3, therefore defines the

X variable used in the expressions on page 309.

Force = 0

Force = 0

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Force Expression

The expression we are going to develop is:

IF({Linear Velocity1}:IF({Linear Displacement1}:IF({Linear

Displacement1}+3:0,0,7.333333*({Linear

Displacement1}+3))+IF({Linear Displacement1}+1.5:0,0,-

7.333333*({Linear Displacement1}+3)-80*({Linear

Displacement1}+3)+131)+IF({Linear

Displacement1}+1.4:0,0,80*({Linear Displacement1}+3)-131-

2.142868*({Linear Displacement1}+3)+6.42857),0,0),0,0)

While this expression, at first, looks complicated, it is just a nested set

of IF statements.

IF Statement The IF statement is used to define an output based on the sign of an

input variable. It is in the form of:

IF (Input variable: A, B, C)

When the value of the Input variable is negative, output the value A.

When the value of the Input variable is zero, output the value B.

When the value of the Input variable is positive, output the value C.

The Input variable, A, B and C can all be either fixed values or

expressions.

In the expression above, we can see that in all the IF statements, there

are only two different input variables, LinearVelocity1 and

LinearDisplacement1.

Developing the Expression

The first thing is to define the point where the blade first touches the

catheter. At this point and before, the force must be zero. From our

measurements, when the blades are open, they are 7.25 mm apart, and

the catheter is 3 mm in diameter. Therefore, the contact occurs when

{LinearDisplacement1}+3=0. We determined this in Step 18.

Therefore, the Force in the first segment will be:

IF({Linear Displacement1}+3:0,0,7.333333*({Linear

Displacement1}+3))

This says,

� When the value of {Linear Displacement1}+3 is negative, the

expression equals zero. � When the value {Linear Displacement1}+3 is zero, then the

expression equals zero.

� When the value of {Linear Displacement1}+3 is positive, then the

value of the expression will be 7.333333 * ({Linear

Displacement1}+3). The value of 7.333333 comes from the

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20 Input the expression.

Edit the force.

Input the above expression. Remember to enter the variable

Linear Displacement1, you can double-click it in the list below the

expression entry box.


Examine the plot of the force.

The plot is correct from distance zero to 5.75 (Point 2 in the graph on

page 308). At that point, the force continues to climb, so we need add

to the IF statement to define Segment 2 (see the graph on page 309).

When you first look at the plot, it may not look correct as we used the

equation of a straight line (7.333333 * ({Linear Displacement1}+3).

Remember however that the linear equation is based on displacement

while the plot is versus time. As the blade motion is not linear, the plot

is correct.

To get through Segment 2 we need to add more to the expression so it

will be:




Displacement1}+3)+131)

If you examine the expression, it is the sum of two IF statements. The

first IF statement is:


Displacement1}+3)

This is the part we already had. We then add to it:

IF({Linear Displacement1}+1.5:0,0,-7.333333*({Linear

Displacement1}+3)-80*({Linear Displacement1}+3)+131)

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This says, when {Linear Displacement1}+1.5 is either negative or

zero, its value will be zero. In other words, until the blade displacement

is 5.75 (Point 2), this part of the expression has no effect.

Once positive, the value will be:

-7.333333*({Linear Displacement1}+3)-80*({Linear

Displacement1}+3)+131

The first part of this is the negative of the first expression, so it is used

to zero the effect of the first expression. The second part of the

expression is the equation for the force in segment 2:

-80*({Linear Displacement1}+3)+131


Edit the force.

Input the above expression.


Examine the plot of the force. For now, we are just interested in the

area circled.

Edit the Y axis so that it shows form -11 to +11. This will make it easier

to see the area of interest.

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The plot is correct from distance zero to 5.85 (Point 3 in the graph on

page 308). At that point, the force needs to reduce at a different rate

based on Segment 3 of our experimental data. So, we will again add to

the IF statement to define Segment 3.

To get through Segment 3 we need to add more to the expression so it

will be:






2.142868*({Linear Displacement1}+3)+6.42857)

Again, the first part of this expression is what we had before. The new

statement is:

IF({Linear Displacement1}+1.4:0,0,80*({Linear Displacement1}+3)-

131-2.142868*({Linear Displacement1}+3)+6.42857)

The first part, IF({Linear Displacement1}+1.4:0,0,80*({Linear

Displacement1}+3)-131is again just the negative of the previous part

of expression to zero it out. The remaining part:

2.142868*({Linear Displacement1}+3)+6.42857)

defines the curve of Segment 3.


Edit the force.




The plot is now correct from zero until the blade finishes the cut

(Point 4). We now have to add another IF statement that will make the

force zero from this point until the end of the blade travel.

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Let’s call the entire expression we have developed thus far Force1.

The IF statement we need is then:

IF ({LinearDisplacement1}: Force1, 0, 0)

When the above expression is negative (blades have not touched yet),

use the entire Force function. If it is zero (the cut is complete) or

negative (blades overlapping), then the force will be zero.

The full expression will now be:

IF({Linear Displacement1}:IF({Linear






2.142868*({Linear Displacement1}+3)+6.42857),0,0)

Let’s call this expression Force2.


Edit the force.



28 Edit the plot.

Change the Y axis back to automatic scaling.

The plot is now correct for the forward travel of the blade, but forces

are mirrored on the blade retraction where they should instead be zero.

To solve this problem, we will add another IF statement, based on the

velocity of the blade that will only use the previously define force

function (Force2) when the blade velocity is negative. That is the

portion of the motion when the surgeon is squeezing the handle and the

blades are closing. When he releases the handle and the spring is

returning the blades to the open position, the blade velocity will be

positive.

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29 Create a new plot.

Create a plot of the Linear Velocity,

X Component of the vertex of the blade shown.

We only want the force to equal the force

function when the velocity is negative. Once the

velocity is zero or positive, it should be zero.

The new IF statement is:

IF({Linear Velocity1}:Force2,0,0)

In the above expression Force2 is used to represent the entire force

function we have already defined. We can see in the expression that it

will only be used when the velocity is negative. When the blades stop

moving and returns, the force will be zero.

If we insert the previous expression for Force2, we get:

IF({Linear Velocity1}:IF({Linear Displacement1}:IF({Linear






2.142868*({Linear Displacement1}+3)+6.42857),0,0),0,0)


Edit the force.


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The plot is now correct for the entire motion of the blade. Its shape is

now the same as the input data from the experiment.

32 Edit the force.

Now that the force

expression is properly

defined, we need to have it

applied to the blades as an

Action/Reaction force.

Change the force from

Action only to Action &

reaction.

Select the two blade

vertices as shown.

Click OK.

33 Modify plot.

Edit the force plot and change it to show the X Component. The

original force was in the Y direction while the direction between the to

blades is the X direction.

Make sure that the force is positive. If it is negative, switch the order of

the vertices in the action/reaction force definition (Step 32).


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Case Study: Surgical Shear - Part 2

In the second part of this Case

Study, we will examine the design

of the handle of the surgical shear.

We have already run the motion

analysis to determine the loads.

Problem Description Determine the stresses on the handle part based on the maximum

loading found in the motion analysis.

Evaluate the handle part for suitability based on the results.


The basic steps are:

� Evaluate the redundancies.

There were several redundancies in the motion study. Each must be

evaluated to determine its effect on the loads needed for the FEA

problem.

� Interference detection.

The assembly must be checked to make sure that parts only contact

each other where designed and that there are no contacts that will

stop the assembly from working properly.

� Export loads.

Exports the loads from SolidWorks Motion to SolidWorks

Simulation.

� Evaluate the imported loads.

The motion loads may not be correct for simulation. Each is

evaluated to make sure it is correct for the FEA process.

� Replace imported loads with local loads.

Loads that were not suitable for FEA must be replaced with the

appropriate load or fixture.

� Run the simulation.

� Evaluate the results.

All results need to be evaluated to make sure the part works as

intended without failure.

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1 Redundancies.

When we were solving the motion simulation, we got several warnings

about the redundancies.

Right-click the local mategroup and click Degrees of Freedom.

There are three redundancies.

Coincident2, has some rotations removed, but that is OK as we are not

concerned with the forces in this mate; it is connecting components

other than the handle.

Concentric1 has both rotations removed. As this is a mate connecting

the handle to an adjacent part it must be checked carefully.

Note that your list of removed degrees of freedom may be slightly

different. This will, however, have no effect on the conclusions

regarding the effect of redundancies made in this step.

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2 Examine the mechanical connection.

The line of action will be through the

center of each part. Because this

connection is not symmetrical, it does not

allow these two forces to act directly on

each other (on the same line of action),

there is a small offset distance which

creates a moment.

This effect is minimized, but not

eliminated, by the offset cut in each of the

two parts.

In the physical model, the two hinge

mates and the concentric mate would all

have some stiffness which would lead to

the redistribution of the torsional moments

between the three connections.

As these torsional moments will be very

small, because they were minimized by

the cutouts, we can ignore this. We will assume that the pins in the two

hinges (between the handle, handle_link and the fixed_cutter) are

very stiff, taking on the torsional loads.

3 Create a plot.

Create a plot of each of the X, Y and Z reaction moment of the hinge

mate (Hinge1) between the handle and the fixed_cutter.

handle_link

handle

Concentric1

TOP View

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The last component of the moment is zero because it is the axial

direction.

The X and Y moments are not zero and have some significant values.

Also note that the moments do not occur at the time when the

maximum cutting force is generated (0.25 seconds), but rather some

time later at about 0.50 seconds.

4 Check interferences.

Before exporting the loads, we need to determine why the large

moment is generated.

Check interference between the latch and moving_cutter. Right-click

the assembly icon in the Motion Study tree and click Check

Interference.

Check from frame 1 to 115 in increments of 1.

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Most of the interferences are very small (volume

<0.01 mm3) and are due to tiny penetrations in the

contact. To locate the individual volumes, select an

interference in the table and click Zoom to

Selection .

If we continue to examine the list of interferences,

we will find some that are several cubic

millimeters. Zoom in on one of these larger

interferences and we will see that at some point,

the latch penetrates the moving_cutter. To fix

this problem, the cutout in the moving cutter has to

be increased in size.

6 Modify the part.

Open the moving_cutter in its own window.

Edit Sketch3 for Cut-Extrude2.

Change the dimensions to increase the size of the slot by 3 mm.

7 Re-check interferences.

Return to the assembly and re-check the interferences.

There should now only be the small contact interferences.


Plot force and moment of hinge connecting the handle to the

fixed_cutter.

9 Locate the maximum force and moment.

Create plots for both the X and Y Force and Moment for the hinge.

Before

After

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We have a high force and moment generated at about 0.14 seconds.

This is not the point where the maximum cutting force is located which

is at 0.24 seconds and shown by the red arrows. You can show the

cutting force plot to verify this location.

10 Examine the latch.

If we examine the latch as the simulation progresses, we can see that

maximum forces and moments are generated when the pin in the latch

goes around the slot path. When the handle is first squeezed, there is a

jump in force as static friction is overcome. As the spring expands

rapidly, the spring works against the forward movement of the

moving_cutter. As the pin gets to the turning point in the slot path, the

forces continue to rise until the pin gets to the horizontal section of the

slot path when the force becomes perpendicular to the path and held by

contact.

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We can see that the maximum force is not caused by cutting the

catheter, but rather by the spring used to retract the mechanism.

11 Determine the frame where

maximum force occurs.

For the plot of the X Component

force, change the X axis scale to

frames.

Change the Start and End values

to 10 and 20 so that we can see

the exact frame.

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12 Moment of the motor.

In the motion simulation, we used a rotary motor to move the

mechanism. When we do the stress analysis, we are going to have to

replace this motor with a force that represents the force applied by the

surgeon on the handle. In order to calculate that force, we need to know

the maximum moment generated by the motor.

Create a plot of the Z Component, Motor Torque of the Rotary Motor. Note the peak torque of 5233 N-mm.

13 Export to FEA.

Click Simulation, Import Motion Loads from the menu.

Note The motion simulation results must be saved before importing the

motion loads.

Export the loads for the handle at frame 15.

14 Open the handle part.

Open the handle in its own window.

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15 Examine loads.

Imported from SolidWorks Motion should be

gravity, a centrifugal load and two remote loads.

The remote load from the handle_link is OK,

however the loads from the rotary motor are not.

When in use, the handle is squeezed by the

surgeon which applies a force directly on the

surface of the handle. Therefore, we will have to

remove the loads from the motor and replace it

with a force.

16 Add a loading force.

We will apply the force at the edge shown

in the image.

Measure the distance from the pivot hole

to the edge. It is approximately 50 mm.

The force we need to apply will be

computed from the torque we measured

in Step 11, 5233/50=104.66 N.

Apply a force of 104.66 N, normal to

Plane5, on edge of handle. Reverse the

direction if necessary to insure the correct

rotation of the handle.

Suppress the remote load from the motor.

Note Your torque value may be slightly different. In such case it is necessary

to update the value of the loading force.

17 Restrain the model.

Suppress the remote load at the pivot point and replace it with a Fixed

Hinge fixture.

18 Add material.

Apply the material Alloy Steel.

19 Mesh.

Mesh the model with a high quality mesh at the default settings.

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We will get a warning:

There is a significant external imbalance force in the Y-direction which will be balanced by the application of opposing inertia forces. Unless your model is under such a force or under marginally imbalance force, application of Inertia Relief may alter the characteristics of your model.

The problem is that we have manually added the force, which is

approximate, so this warning is expected.

Click Yes.

We will get another warning:

Excessive displacements were calculated in this model. If your system is properly restrained, consider using the Large Displacement option to improve the accuracy of the calculations. Otherwise continue with the current settings and review the causes of these displacements.

Since the external forces are in slight imbalance, the handle wants to

rotate about the hinge support. This is an unavoidable consequence

having no implication on the resulting stresses and deformations. The

handle will rotate as a rigid body.

Click No to continue with the linear solution.


The maximum stress about 252 MPa and is located at the sharp edge

under the pivot.

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A closer examination shows that the highest

stress is at the connection of the cylindrical

sections to the pivot. A more detail analysis

of this area might be appropriate since this is

a stress singularity location.

Given the yield strength of 620 MPa, the

stresses in the handle are acceptable.

22 Create a Factor of Safety plot.

The plot shows that the Factor of Safety is

close to 2.5, so the design is acceptable.

Note The above plot has the upper limit set to 100.


Summary In this lesson, we analyzed a surgical shear assembly utilized to cut a

catheter. Since a catheter is not a rigid object, its resistance against the

cut is expressed by a reaction force acting on the blades. In rigid body

dynamics software, we cannot simulate the deformation and separation

of the flexible catheter and, therefore, modeling of an equivalent

catheter reaction force is required.

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Real experimental data for a catheter cutting force was used to

construct a complex, position dependent, expression for the equivalent

action/reaction force. Due to the complexity of the problem, the

expression involved multiple uses of the IF statement. Various graphs

were generated at the end of each analysis. Motion mate forces were

then imported into SolidWorks Simulation for the finite element stress

analysis. The analysis indicated that the handle component is designed

with a satisfactory safety factor of 2.5.

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Appendix AMotion Study Convergence

Solutions and AdvancedOptions

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Appendix A SolidWorks 2011Motion Study Convergence Solutions and Advanced Options

334

Convergence Complex assemblies with many redundancies or problems featuring

scenarios numerically difficult to overcome (for example instability

points featured in Lesson 4, fast changing motions or high velocity

impacts to name a few) may cause the solver to fail to converge and the

solution may terminate before reaching the end. Convergence issues

are unavoidable consequences of the numerical simulation and certain

expertise may be required to overcome them. On occasions we may

foresee that a complex assembly will pose difficulty and will need more

attention. In general, however, it is difficult to predict when the

convergence issues may occur. The following text should help you

understand the basics of solving the above mentioned difficulties. It

introduces some of the advanced software features not used during the

regular part of the course.

When the SolidWorks Motion solver faces convergence problems, the

motion study terminates and the following window opens:

There are a few possible reasons for the convergence issues; we will

review them in the next paragraphs.

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335

Accuracy A set of coupled differential and algebraic equations

(DAE) define the equations of motion of a SolidWorks Motion model. A solution to the equations of motion is obtained by solving these equations using an integrator. The integrator obtains the solution in two stages: first it predicts the solution at the next time step based on the past history and then it corrects that solution based on the state data at that time until the solution is within the desired accuracy level.

The Accuracy setting controls how accurate you want your solution to be. There is a trade-off between accuracy and performance. If the Accuracy setting is towards the High end, then the motion solver may take a long time to compute the solution. On the other hand, if this setting is towards the Low end, then the results may not be very accurate.

While the default value of 0.0001 fits most situations; it may need to be changed if sudden changes in the system occur. In such situations, the predictor provides an incorrect initial guess to the corrector resulting in large error and failure. This value may need to be reduced when sudden and discontinuous changes occur during simulation; such as sudden changes in the force or motor magnitudes, use of non-differentiable intrinsic functions in the statements (IF, MIN, MAX, SIGN, MOD, and DIM), friction and similar.

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The advanced options listed below can be accessed by clicking the

Advanced Options in the Motion Study Properties shown in the figure

above.

Integrator Type The SolidWorks Motion solver solves the DAE equations of motion by

integrating the differential equations in such a way that the algebraic

constraint equations are also satisfied at every time step. The speed of

the solution depends upon the numerical stiffness of these equations;

the stiffer the equations the slower the solution. A set of ordinary

differential equations are characterized as numerically stiff when there

is a wide spread between high and low frequency eigenvalues, with the

high-frequency eigenvalues being overdamped. Special efficient

integration methods are required to solve numerically stiff differential

equations because usual methods for solving differential equations

perform poorly and are too slow.

The SolidWorks Motion solver offers three stiff integration methods for

computing motion:

� GSTIFF

� WSTIFF

� SI2

GSTIFF The GSTIFF integration method developed by C. W. Gear is a variable

order, variable step size integration method. It is the default method

used by the SolidWorks Motion solver. The GSTIFF method is a fast

and accurate method for computing displacements for a wide range of

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WSTIFF WSTIFF is another variable order, variable step size stiff integrator.

Both methods are very similar in formulation and behavior. Both of

them use a backwards difference formulation. The only difference is

that the coefficients used internally by GSTIFF are calculated assuming

a constant step size whereas in WSTIFF, these coefficients are a

function of the step size. If the step size changes suddenly during

integration, GSTIFF introduces a small error in the solution whereas

WSTIFF can handle it without any loss of accuracy. So the problems

run more smoothly in WSTIFF. Sudden step size changes occur

whenever there are discontinuous forces, discontinuous motions or

abrupt events such as 3D contacts in the model.

Stabilized Index Two (SI2)

The Stabilized Index Two (SI2) method offered in SolidWorks Motion

is a modification of the GSTIFF integration method. This method

provides better error control over velocity and acceleration terms in the

equations of motion. Provided the motion is sufficiently smooth, SI2

velocity and acceleration results are more accurate than those

computed with GSTIFF or WSTIFF, even for motions with high

frequency oscillations. SI2 is also more accurate with smaller step

sizes, but is significantly slower.

Integrator Settings

With each integrator, there are several settings that control the step size

and number of integration steps.

Maximum Iterations

Maximum Iterations parameter controls the maximum number of

iterations the SolidWorks Motion solver may use to converge to a

solution. The default value of 25 iterations should suffice for most

problems. It is not recommended to increase this parameter

substantially; this parameter is typically not the cause of the failure.

Initial Integrator Step Size

Initial Integrator Step Size controls the value of the step at the first

solution instance. If your simulation faces some difficulties at the initial

stages of the solution, consider reducing this value. Typically, this

parameter does not need to be changed.

Minimum Integrator Step Size

During the integration process, if the simulation error is too large the

integrator reduces the time step and attempts the solution again until

the desired accuracy is satisfied. The integrator will not reduce the step

size beyond the Minimum Integrator Step Size. The default magnitude

is acceptable for most of the simulation and does not need to be

changed.

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Maximum Integrator Step Size

The Maximum Integrator Step Size controls the value of the largest

time step the integrator may take during the solution. Increasing the

Maximum Integrator Step Size speeds up the solution, and reduces the

time required to solve the model. But if this value is too big, there is a

chance the solver may take too large a step and enter a region from

which it may not recover and hence fail to converge. Reducing this

value has no effect on the accuracy of the solution. When using

GSTIFF integrator, velocities and accelerations may have

discontinuities for larger values of the integrator time step. You can

reduce this error by reducing the maximum integrator step size. If you

know that the motion is smooth and there are no such abrupt changes,

you can increase this value to speed up the solution. When facing

convergence problems, modifying this parameter may help.

If there are abrupt changes in forces or motions happening over small

time durations that you may need to reduce the maximum integrator

step size to make sure that the integrator does not miss such events.

You may want to reduce this value if you have contact between a solid

body and a thin body, and the solver fails to recognize this contact. This

can happen, for example, if you have a ball bouncing on a thin plate.

Depending upon your model parameters, it is possible that the ball may

pass through the plate without recognizing the contact between them.

In such a case, reducing the Maximum Integrator Step Size forces the

solver to take smaller steps so that it does not miss the contact

incidence between the two bodies.

Reducing this value will slow down the integrator but it has no effect

on the accuracy of the results. On the other hand, if you know that the

motion is smooth and there are no such abrupt changes, you can

increase this value to speed up the solution.

Jacobian Re-evaluation

The Jacobian Matrix is a matrix of partial derivatives required to solve

the linearized approximation of the original nonlinear equations of

motion during the Newton-Raphson iteration procedure. Users may

find it useful to view this matrix similar to what the stiffness matrix is

in the finite element analysis. The default setting, the most accurate and

also the most time consuming is the re-evaluation of the Jacobian

Matrix at every iteration. While reducing the re-evaluation speeds up

the solution, it should only be done when changes in the assembly

motion are slow. Setting of this parameter has no effect on the accuracy,

but too low a setting may cause the integrator to fail.

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Conclusion The most common parameters that need to be adjusted when you face

convergence difficulties are Accuracy, Maximum Integrator Step Size,

and Contact Resolution. If chancing none of the above parameters help

the convergence, make sure that your inputs are smooth and

differentiable, especially the user input expressions with mathematical

functions. It is advisable to use STEP function rather than IF statement.

On occasions redundant constraint may cause the integrator to fail

because the solver is having difficulty satisfying the constraints. The

most likely cause for such a failure is an inconsistently defined or ill-

behaved model. In these situations try to eliminate the redundancy or

the mating relationships in the assembly.

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Appendix BMate Friction

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Appendix B SolidWorks 2011Mate Friction

342

Mate Friction Friction is a force that occurs in mates and parts in contact. When parts

are in contact, friction is calculated based on the static and dynamic

coefficients of friction and the normal force acting on the part. Mate

friction is more complex, because the size of the mate can affect the

magnitude of friction.

In 1699, Amontons rediscovered Leonardo da Vinci's two laws of

friction: the frictional force is directly proportional to the normal load,

and the size of the bodies does not affect the friction [Bowden and

Tabor, 1950, 1974]. Engineers have relied on Amontons' laws,

extensively and routinely for three centuries. Contrary to popular

belief, the size of the bodies does affect the friction forces in the case of

mate friction.

Mate friction is a resistive, sliding, surface force between parts that

must be overcome for the parts to move with respect to one another.

The force develops due to contact between the surfaces and the loads

acting on the connection. For a pin in a hole, mate friction is

experienced as an additional torque restricting the pin from rotating

with respect to the hole. Mate friction is not anything more than

standard friction between bodies; however it takes into account aspects

of components’ geometries in determining the net frictional forces

acting.

For example, think of a pin in a

hole, but with a little slope. In

the first image shown below,

the pin is resting in the hole

under a centrally located force.

This is the equivalent of a pure

bearing load. The force needed

to slide the pin back and forth is only dependent on the vertical load.

The torque needed to rotate the pin is dependent on this force, but also

on the radius of the pin (see second image below). In this example, the

radius of the pin has no effect on the magnitude of the friction force,

but does have an effect on the moment required to overcome friction to

rotate the pin (mu.r.F)

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343

Now, consider the case where there is

an additional moment on the pin. The

moment forces the pin to rotate,

becoming supported at the outer edges

of the hole (w). The moment is reacted

as a force couple (M/w). Dividing the

bearing load (F) between the ends,

results in local force of F/2 + M/w.

Frictional forces are accumulative so you can sum these force couples

to get the total force upon which friction is based (F+2M/w).

It is a simple extension of this to derive the torque necessary to rotate

the pin as mu*r(F+2M/w).

This influence of the bending moment of a mate is an important factor

in mate friction. You can see that if the hole supporting the pin is not

thick (in terms of w), the moment component tends to be very high. If

the hole supporting the pin is very thick, the moment component tends

towards zero.

Concentric, coincident and many other SolidWorks mates support fric-tion. When friction effects are enabled for these mates, a force is induced that opposes the motion of the mates and is a function of the reaction forces acting on the mate.

Where to Find It � In the SolidWorks mate property manager, Analysis tab, Friction

dialog.

Concentric (Spherical) Mate Friction Model

For the purpose of calculating friction effects, a

Concentric (Spherical) mate is modeled as a

ball rotating in a socket. Some portion of the

ball's surface area is in contact with the socket.

Dimension d is the diameter of the ball.

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Coincident Translational Mate Friction Model


Coincident (translational) mate is modeled as a

rectangular bar sliding in a rectangular sleeve.

Dimension h is the height of the rectangular bar.

Dimension w is the width of the rectangular bar.

Dimension l is the length of the bar that is in

contact with the sleeve.

Concentric Mate Friction Model


Concentric mate is modeled as a snug fit pin

rotating and sliding in a hole. Dimension r is the

radius of the pin, and Dimension l is the length of

the pin that is in contact with the hole.

Concentric mate friction model can only be

activated for faces. No edges are allowed.

Coincident Mate (Planar) Friction Model

For the purpose of calculating friction effects, this

is modeled as one block sliding and rotating across

the surface of another block. Dimensions l and w

are the length and width of the sliding block.

Dimension r is the radius of a circle, centered at

the center of the block face that circumscribes the

face of the sliding block that is in contact with the

other block.

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SolidWorks 2011 Appendix BMate Friction

345

Universal Joint Friction Model


universal joint is modeled as a cylindrical cross

piece rotating in a set of end caps. Dimension r is

the radius of the bearing end cap. Dimension w is

the height of the cross pieces.

Friction Results Reported

Joint Type Friction Force Friction Moment

Concentric (two faces) Yes Yes

Concentric (two spheres) No Yes

Universal No Yes

Coincident (translational) Yes No

Coincident (planar) Yes Yes

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347

Index

A

accuracy 335action 294action and reaction forces 17action forces 17applied forces 18

B

bushingsdefining 204properties 224

C

CAM 166desmodromic 179profile 167rocker 185

chart properties 119closing force 123coefficient of restitution 110constraint forces 8constraint mapping 8contact 76, 82

curve to curve 155, 159contact

forces 154

forces 100, 280friction 84precise contact 123precise geometry 115solid bodies 159tessellated geometry 115

contact, solid bodies - contact 109convergence 334Cycle Based Motion 173

D

dampertranslational 87

dampingcoefficient c 111penetration d 111

degrees of freedom 219calculation 219estimated 219total actual 219

dynamic systems 212, 229, 231

E

event based motion view 294event based simulation 290export

results to FEA 262trace path curves 170

F

fixed parts 7flexible joints 194, 204

bushings 204flexible mates

limitations 223floating parts 8force expression 313force function 105, 308forces 17

action and reaction 17action only 17applied 18closing 123contact 100, 154, 280definition 18impact 110

frames per second 26friction 76, 342

contact 84kinematic coefficient 84static coefficient 84

Function Builder 45

G

gravity 8, 16GSTIFF 124, 336

I

IF statement 313impact force 110

exponent e 111stiffness k 111

instability 118integrator settings 337integrator types 123, 336

GSTIFF 124, 336SI2 124, 337WSTIFF 124, 337

interference detection 81

J

jacobian 338

Jerk 46joints

flexible 194, 204rigid 194

K

kinematic coefficient of friction 84kinematic systems 212, 233

L

linear springmagnitude - spring force 86

load bearing faces 263

M

mass properties 131mate friction 342

concentric 344planar 344results 345spherical 343translational 344universal 345

mates 8, 342modifying plots 119motion

driving 14, 45, 173motion study properties 26

frames per second 26precise contact 123

motorservo motor 290

motors 8, 14fixing motion 101force function 105force types 103functional expressions 103

P

Path Mate 144PathMate 144plots

modifying 119resizing 21

Plotting Kinematic Results 54plotting results 21Poisson model 110postprocessing 89precise contact 123precise geometry 115

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Index SolidWorks 2011

348

propertiesbushings 224chart 119mass 131motion study 26

proximity sensor 291

R

redundanciescheck 225defined 213effects 214mechanisms 226removal by solver 215removal with bushings 244removal with flexible joints 222removed in the solver 215zero 238, 243

redundant mechanisms 226resizing plots 21restitution coefficient 110results

export 258export to FEA 262plotting 21

rigid body 7rigid joints 194

S

sensors 291proximity 291

servo motor 290SI2 337SI2 (Stabilized Index Two) 124spring

force magnitude 86translational 85

spring force 86static coefficient of friction 84STEP function 105

T

task 293, 295action 294triggers 293

tessellated geometry 115timeline view 294toe angle 201trace path 169

export curves 170translational damper 87translational spring 85triggers 293

V

viewevent based 294timeline 294

W-Z

WSTIFF 124, 337

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Documents

SolidWorks Motion Tutorials